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# Solved Examples on Trigonometric Expressions

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Simplify the trigonometric expressions completely
State the reasons for every step
Show all work
There should be no negative angles

(1.) ACT For all $x$ such that $0 \le x \le 90$, which of the following expressions is NOT equal to $\sin x^\circ$?

$F.\:\: -\sin(-x^\circ) \\[3ex] G.\:\: \sin(-x^\circ) \\[3ex] H.\:\: \cos(90 - x)^\circ \\[3ex] J.\:\: \cos(x - 90)^\circ \\[3ex] K.\:\: \sqrt{1 - (\cos x^\circ)^2}$

$F.\:\: -\sin(-x^\circ) = -1 * \sin(-x) \\[3ex] \sin(-x) = -\sin x ...Odd\:\: Identity \\[3ex] = -1 * -\sin x = \sin x ...YES \\[5ex] G.\:\: \sin(-x^\circ) \\[3ex] \sin(-x) = -\sin x ...Odd\:\: Identity \\[3ex] -\sin x \ne \sin x ...NO \\[5ex] H.\:\: \cos(90 - x)^\circ \\[3ex] \cos(90 - x) = \sin x ...Cofunction\:\: Identity \\[3ex] YES \\[5ex] J.\:\: \cos(x - 90)^\circ \\[3ex] -(x - 90) = -x + 90 = 90 - x \\[3ex] \cos(x) = \cos(-x) ...Even\:\: Identity \\[3ex] \cos(x - 90) = \cos(90 - x) ...Even\:\: Identity \\[3ex] = \sin x ...YES \\[5ex] K.\:\: \sqrt{1 - (\cos x^\circ)^2} \\[3ex] (\cos x)^2 = \cos^2 x \\[3ex] 1 - \cos^2 x = \sin^2 x ...Pythagorean\:\: Identity \\[3ex] = \sqrt{\sin^2 x} = \sin x ...YES$
(2.) ACT The expression $(180 - x)$ is the degree measure of a nonzero acute angle if and only if:

$A.\;\; 0 \lt x \lt 45 \\[3ex] B.\;\; 0 \lt x \lt 90 \\[3ex] C.\;\; 45 \lt x \lt 90 \\[3ex] D.\;\; 90 \lt x \lt 135 \\[3ex] E.\;\; 90 \lt x \lt 180 \\[3ex]$

Second Quadrant Identities or Identities of Supplements

$90 \lt \theta \lt 180$ ... Angle in Degrees
Reference Angle = $180 - \theta$ ... Angle in Degrees

The expression $(180 - x)$ is the degree measure of a nonzero acute angle if and only if: $90 \lt x \lt 180$
(3.) $(\sin \theta + \csc \theta)(\sin \theta - \csc \theta)$

$(\sin \theta + \csc \theta)(\sin \theta - \csc \theta) \\[3ex] = \sin^2 \theta - \csc^2 \theta ... Difference\:\: of\:\: Two\:\: Squares$
(4.) $\cos \theta \sin \theta(\tan \theta + \csc \theta)$

$\cos \theta \sin \theta(\tan \theta + \csc \theta) \\[3ex] = \cos \theta \sin \theta \tan \theta + \cos \theta \sin \theta \csc \theta \\[3ex] But;\:\: \tan \theta = \dfrac{\sin \theta}{\cos \theta} ... \:Quotient \:Identity \\[5ex] \csc \theta = \dfrac{1}{\sin \theta} ... \:Reciprocal \:Identity \\[5ex] Substitute \:these\: identities \\[3ex] = \cos \theta \sin \theta * \dfrac{\sin \theta}{\cos \theta} + \cos \theta \sin \theta * \dfrac{1}{\sin \theta} \\[5ex] = \sin \theta \sin \theta + \cos \theta \\[3ex] = \sin^2 \theta + \cos \theta$
(5.) $\dfrac{1}{1 + \cos \theta} + \dfrac{1}{1 - \cos \theta}$

$\dfrac{1}{1 + \cos \theta} + \dfrac{1}{1 - \cos \theta} \\[5ex] LCD = (1 + \cos \theta)(1 - \cos \theta) \\[5ex] = \dfrac{1(1 - \cos \theta) + 1(1 + \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)} \\[5ex] = \dfrac{1 - \cos \theta + 1 + \cos \theta}{(1 + \cos \theta)(1 - \cos \theta)} \\[5ex] = \dfrac{2}{(1 + \cos \theta)(1 - \cos \theta)} \\[5ex] (1 + \cos \theta)(1 - \cos \theta) = 1^2 - \cos^2 \theta = 1 - \cos^2 \theta ...\: Difference\: of\: Two\: Squares \\[3ex] = \dfrac{2}{1 - \cos^2 \theta} \\[5ex] 1 - \cos^2 \theta = \sin^2 \theta ...\:Pythagorean\: Identity \\[3ex] = \dfrac{2}{\sin^2 \theta} \\[5ex] \dfrac{1}{\sin \theta} = \csc \theta ...\:Reciprocal\: Identity \\[5ex] = 2 * \dfrac{1}{\sin \theta} * \dfrac{1}{\sin \theta} \\[5ex] = 2 * \csc \theta * \csc \theta \\[3ex] = 2\csc^2 \theta$
(6.) $\dfrac{3\sin^2 \theta + \sin \theta - 10}{\sin \theta + 2}$

$Let\: p = \sin \theta \\[3ex] \dfrac{3\sin^2 \theta + \sin \theta - 10}{\sin \theta + 2} \\[5ex] = \dfrac{3p^2 + p - 10}{p + 2} \\[5ex] = \dfrac{(p + 2)(3p - 5)}{p + 2} \\[5ex] = 3p - 5 \\[3ex] = 3\sin \theta - 5$
(7.) $\dfrac{\cos \theta}{1 + \csc \theta} * \dfrac{1 - \csc \theta}{1 - \csc \theta}$

$\dfrac{\cos \theta}{1 + \csc \theta} * \dfrac{1 - \csc \theta}{1 - \csc \theta} \\[5ex] = \dfrac{\cos \theta(1 - \csc \theta)}{(1 + \csc \theta)(1 - \csc \theta)} \\[5ex] (1 + \csc \theta)(1 - \csc \theta) = 1^2 - \csc^2 \theta ... \:Difference\: of \:two \:Squares \\[3ex] = \dfrac{\cos \theta - \cos \theta \csc \theta}{1^2 - \csc^2 \theta} \\[5ex] 1^2 - \csc^2 \theta = 1 - \csc^2 \theta \\[3ex] 1 - \csc^2 \theta = -\cot^2 \theta ... \:Pythagorean\: Identity \\[3ex] = \dfrac{\cos \theta - \cos \theta \csc \theta}{-\cot^2 \theta} \\[5ex] \csc \theta = \dfrac{1}{\sin \theta} ... \:Reciprocal\: Identity \\[5ex] \tan \theta = \dfrac{1}{\cot \theta} ... \:From \:Reciprocal\: Identity \\[5ex] \cot \theta = \dfrac{\cos \theta}{\sin \theta} ... \:Quotient\: Identity \\[5ex] = \dfrac{\cos \theta - \cos \theta * \dfrac{1}{\sin \theta}}{-\cot^2 \theta} \\[5ex] = \dfrac{\cos \theta - {\dfrac{\cos \theta}{\sin \theta}}}{-\cot^2 \theta} \\[5ex] = \dfrac{\cos \theta - \cot \theta}{-\cot^2 \theta} \\[5ex] = \dfrac{\cos \theta}{-\cot^2 \theta} - \dfrac{\cot \theta}{-\cot^2 \theta} \\[5ex] = \dfrac{-\cos \theta}{\cot^2 \theta} - \dfrac{\cot \theta}{-\cot^2 \theta} \\[5ex] = (-\cos \theta \div \cot^2 \theta) + \dfrac{1}{\cot \theta} \\[5ex] = (-\cos \theta \div \dfrac{\cos^2 \theta}{\sin^2 \theta}) + \tan \theta \\[5ex] = (-\cos \theta * \dfrac{\sin^2 \theta}{\cos^2 \theta}) + \tan \theta \\[5ex] = \dfrac{-\sin^2 \theta}{\cos \theta} + \tan \theta \\[5ex] = -\sin \theta * \dfrac{\sin \theta}{\cos \theta} + \tan \theta \\[5ex] = -\sin \theta \tan \theta + \tan \theta \\[3ex] = \tan \theta - \sin \theta \tan \theta \\[3ex] = \tan \theta(1 - \sin \theta)$
(8.) $\dfrac{\sin^2 \theta - 25}{2\cos \theta + 1} * \dfrac{4\cos \theta + 2}{5\sin \theta + 25}$

$\dfrac{\sin^2 \theta - 25}{2\cos \theta + 1} * \dfrac{4\cos \theta + 2}{5\sin \theta + 25} \\[5ex] \sin^2 \theta - 25 = \sin^2 \theta - 5^2 \\[3ex] \sin^2 \theta - 5^2 = (\sin \theta + 5)(\sin \theta - 5) ... \:Difference \:of \:Two \:Squares \\[3ex] 4\cos \theta + 2 = 2(2\cos \theta + 1) \\[3ex] 5\sin \theta + 25 = 5(\sin \theta + 5) \\[3ex] = \dfrac{(\sin \theta + 5)(\sin \theta - 5)}{2\cos \theta + 1} * \dfrac{2(2\cos \theta + 1)}{5(\sin \theta + 5)} \\[5ex] = \dfrac{1 * (\sin \theta - 5)}{1} * \dfrac{2 * 1}{5 * 1} \\[5ex] = \dfrac{2(\sin \theta - 5)}{5}$
(9.) ACT Which of the following is equivalent to $\dfrac{1 - \cos^2 \theta}{\cos^2 \theta}$

$F.\:\: \sec^2 \theta \\[3ex] G.\:\: (\csc^2 \theta) - 1 \\[3ex] H.\:\: \tan^2 \theta \\[3ex] J.\:\: \sin^2 \theta \\[3ex] K.\:\: -\dfrac{1}{\sin^2 \theta}$

We can solve this question in two ways.
Use whatever way you prefer

$\underline{First\:\: Method} \\[3ex] \dfrac{1 - \cos^2 \theta}{\cos^2 \theta} \\[5ex] \sin^2 \theta + \cos^2 \theta = 1 ...Pythagorean\:\: Identity \\[3ex] \implies \sin^2 \theta = 1 - \cos^2 \theta \\[3ex] = \dfrac{\sin^2 \theta}{\cos^2 \theta} \\[5ex] \tan \theta = \dfrac{\sin \theta}{\cos \theta} \\[5ex] = \left(\dfrac{\sin \theta}{\cos \theta}\right)^2 \\[5ex] = \tan^2 \theta \\[3ex] \underline{Second\:\: Method} \\[3ex] \dfrac{1 - \cos^2 \theta}{\cos^2 \theta} \\[5ex] = \dfrac{1}{\cos^2\theta} - \dfrac{\cos^2\theta}{\cos^2\theta} \\[5ex] \dfrac{1}{\cos\theta} = \sec\theta \\[5ex] \rightarrow \dfrac{1}{\cos^2\theta} = \sec^2\theta \\[5ex] = \sec^2\theta - 1 \\[3ex] 1 + \tan^2\theta = \sec^2\theta ... Pythagorean\:\: Identity \\[3ex] \tan^2\theta = \sec^2\theta - 1 \\[3ex] = \tan^2\theta$
(10.) $\dfrac{\sin \theta}{1 - \cos \theta} * \dfrac{1 + \cos \theta}{1 + \cos \theta}$

$\dfrac{\sin \theta}{1 - \cos \theta} * \dfrac{1 + \cos \theta}{1 + \cos \theta} \\[5ex] = \dfrac{\sin \theta(1 + \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} \\[5ex] (1 + \cos \theta)(1 - \cos \theta) = 1^2 - \cos^2 \theta ... \:Difference\: of \:two \:Squares \\[3ex] 1^2 - \cos^2 \theta = 1 - \cos^2 \theta \\[3ex] 1 - \cos^2 \theta = \sin^2 \theta ... \:Pythagorean\: Identity \\[3ex] = \dfrac{\sin \theta(1 + \cos \theta)}{\sin^2 \theta} \\[5ex] = \dfrac{1 + \cos \theta}{\sin \theta} \\[5ex] = \dfrac{1}{\sin \theta} + \dfrac{\cos \theta}{\sin \theta} \\[5ex] \csc \theta = \dfrac{1}{\sin \theta} ... \:Reciprocal\: Identity \\[5ex] \cot \theta = \dfrac{\cos \theta}{\sin \theta} ... \:Quotient\: Identity \\[5ex] = \csc \theta + \cot \theta$
(11.) $2\sin^3\omega\cos\omega + 2\sin\omega\cos^3\omega$

$2\sin^3\omega\cos\omega + 2\sin\omega\cos^3\omega \\[3ex] = 2\sin\omega\cos\omega(\sin^2\omega + \cos^2\omega) \\[3ex] \sin^2\omega + \cos^2\omega ... Pythagorean\:\: Identity \\[3ex] 2\sin\omega\cos\omega = \cos2\omega ... Double-Angle\:\: Formula \\[3ex] = \cos2\omega * 1 \\[3ex] = \cos2\omega$
(12.) $\dfrac{(\cos \theta - \sin \theta)(\cos \theta - \sin \theta) - 1}{\sin \theta \cos \theta}$

$\dfrac{(\cos \theta - \sin \theta)(\cos \theta - \sin \theta) - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{(\cos^2 \theta - \sin \theta \cos \theta - \sin \theta \cos \theta + \sin^2 \theta) - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{(\cos^2 \theta - 2\sin \theta \cos \theta + \sin^2 \theta) - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{(\cos^2 \theta + \sin^2 \theta - 2\sin \theta \cos \theta) - 1}{\sin \theta \cos \theta} \\[5ex] \cos^2 \theta + \sin^2 \theta = 1 ... \:Pythagorean\: Identity \\[3ex] = \dfrac{(1 - 2\sin \theta \cos \theta) - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{1 - 2\sin \theta \cos \theta - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{- 2\sin \theta \cos \theta}{\sin \theta \cos \theta} \\[5ex] = -2$
(13.) $\dfrac{8\cot \theta \csc \theta - 2\csc \theta}{4\cot \theta \csc \theta + 2\csc \theta}$

$\dfrac{8\cot \theta \csc \theta - 2\csc \theta}{4\cot \theta \csc \theta + 2\csc \theta} \\[5ex] = \dfrac{2\csc \theta(4\cot \theta - 1)}{2\csc \theta(2\cot \theta + 1)} \\[5ex] = \dfrac{4\cot \theta - 1}{2\cot \theta + 1}$
(14.) ACT The expression $4 \sin x \cos x$ is equivalent to which of the following?
(Note: $\sin(x + y) = \sin x \cos y + \cos x \sin y$)

$F.\:\: 2\sin 2x \\[3ex] G.\:\: 2\cos 2x \\[3ex] H.\:\: 2\sin 4x \\[3ex] J.\:\: 8\sin 2x \\[3ex] K.\:\: 8\cos 2x$

$4\sin x \cos x = 2\sin x \cos x + 2\sin x \cos x \\[3ex] = 2(\sin x \cos x + \sin x \cos x) \\[3ex] \sin(x + x) = \sin x \cos x + \cos x \sin x = \sin x \cos x + \sin x \cos x \\[3ex] = 2\sin(x + x) \\[3ex] = 2\sin 2x$
(15.) ACT Whenever $\dfrac{\tan \theta}{\sin \theta}$ is defined, it is equivalent to:

$F.\:\: \cos \theta \\[3ex] G.\:\: \dfrac{1}{\cos \theta} \\[5ex] H.\:\: \dfrac{1}{\sin \theta} \\[5ex] J.\:\: \dfrac{1}{\sin^2 \theta} \\[5ex] K.\:\: \dfrac{\cos \theta}{\sin^2 \theta}$

$\dfrac{\tan \theta}{\sin \theta} \\[5ex] = \tan \theta \div \sin \theta \\[3ex] = \tan \theta * \dfrac{1}{\sin \theta} \\[5ex] \tan \theta = \dfrac{\sin \theta}{\cos \theta} ...Reciprocal\:\: Identity \\[5ex] = \dfrac{\sin \theta}{\cos \theta} * \dfrac{1}{\sin \theta} \\[5ex] = \dfrac{1}{\cos \theta} \\[5ex] = \sec \theta$
(16.) $\sin \theta(\sin \theta - 2\cos \theta) + \cos^2 \theta$

$\sin \theta(\sin \theta - 2\cos \theta) + \cos^2 \theta \\[3ex] = \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta \\[3ex] = \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta \\[3ex] \sin^2 \theta + \cos^2 \theta = 1 ... \:Pythagorean\: Identity \\[3ex] 2\sin \theta \cos \theta = \sin 2\theta ... \:Double-Angle\: Formulas \\[3ex] = 1 - \sin 2\theta$
(17.) $(\cos \theta + \sec \theta)(3\cos^2 \theta + 3\sec^2 \theta - 3)$

$(\cos \theta + \sec \theta)(3\cos^2 \theta + 3\sec^2 \theta - 3) \\[3ex] = 3\cos^3 \theta + 3\cos \theta \sec^2 \theta - 3\cos \theta + 3\cos^2 \theta \sec \theta + 3\sec^3\theta -3\sec\theta \\[3ex] \sec\theta = \dfrac{1}{\cos\theta} ... Reciprocal\:\: Identity \\[3ex] 3\cos\theta\ sec^2\theta = 3 * \cos\theta * \sec\theta * \sec\theta = 3 * \cos\theta * \dfrac{1}{\cos\theta} * \sec\theta = 3\sec\theta \\[5ex] 3\cos^2 \theta \sec \theta = 3 * \cos\theta * \cos\theta * \dfrac{1}{\cos\theta} = 3\cos\theta \\[5ex] = 3\cos^3 \theta + 3\sec\theta - 3\cos \theta + 3\cos\theta + 3\sec^3\theta -3\sec\theta \\[3ex] = 3\cos^3 \theta + 3\sec^3\theta + 3\sec\theta - 3\sec\theta - 3\cos \theta + 3\cos\theta \\[3ex] = 3\cos^3\theta + 3\sec^3\theta$
(18.) $\tan^3 \theta + 27$

$\tan^3 \theta + 27 \\[3ex] = \tan^3 \theta + 3^3 \\[3ex] x^3 +y^3 = (x + y)(x^2 - xy + y^2) ...\:Sum\: of\: Two\: Cubes \\[3ex] Compare\: and\: Substitute \\[3ex] x = \tan \theta \\[3ex] y = 3 \\[3ex] = (\tan \theta + 3)(\tan^2 \theta - 3\tan \theta + 3^2) \\[3ex] = (\tan \theta + 3)(\tan^2 \theta - 3\tan \theta + 9)$
(19.) Rationalize the numerator

$\sqrt{\dfrac{7 + 7\cos \theta}{7 - 7\cos \theta}}$

We want to rationalize the numerator
This means that we want to make the numerator a rational number
What do we have to multiply by $1$ (the exponent at the denominator is $1$) in terms of exponent so that it is divisible by $2$
Why $2$? ... it is because of the square root.
Remember: Square root means an exponent of $\dfrac{1}{2}$... division by $2$
$\sqrt{x} = x^\frac{1}{2}$
So, we have to multiply $7 + 7\cos \theta$ by $7 + 7\cos \theta$ so we can get the squared of it $(7 + 7\cos \theta)^2$ which is divisible by $2$
Whatever you do to the numerator, you have to do the same thing to the denominator
This is necessary in order to get an equivalent fraction to the original fraction
$\sqrt{\dfrac{7 + 7\cos \theta}{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} * \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 + 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{(7 + 7\cos \theta) * (7 + 7\cos \theta)}}{\sqrt{(7 - 7\cos \theta)* (7 + 7\cos \theta)}} \\[5ex] (7 - 7\cos \theta) * (7 + 7\cos \theta) = 7^2 - (7\cos \theta)^2 ...\:Difference\: of\: Two\: Squares \\[3ex] (7 - 7\cos \theta) * (7 + 7\cos \theta) = 49 - 49\cos^2 \theta \\[3ex] = \dfrac{\sqrt{(7 + 7\cos \theta)^2}}{\sqrt{49 - 49\cos^2 \theta}} \\[5ex] = \dfrac{7 + 7\cos \theta}{\sqrt{49(1 - \cos^2 \theta)}} \\[5ex] 1 - \cos^2 \theta = \sin^2 \theta ...\:Pythagorean\: Identity \\[3ex] = \dfrac{7(1 + \cos \theta)}{\sqrt{49\sin^2 \theta}} \\[5ex] = \dfrac{7(1 + \cos \theta)}{\sqrt{49} * \sqrt{\sin^2 \theta}} \\[5ex] = \dfrac{7(1 + \cos \theta)}{7\sin \theta} \\[5ex] = \dfrac{1 + \cos \theta}{\sin \theta} \\[5ex]$ The numerator is now a rational number.
(20.) $12\cos^2\alpha + \cos\alpha - 13$

This is a quadratic expression in $\cos\alpha$
Let us make this as simple as possible.
Let us use substitution
Then, try to use the Factoring method
Remember to substitute back

$12\cos^2\alpha + \cos\alpha - 13 \\[3ex] Let\:\: \cos\alpha = p \\[3ex] 12p^2 + p - 13 \\[3ex] 12p^2 + 13p - 12p - 13 \\[3ex] p(12p + 13) - 1(12p + 13) \\[3ex] (12p + 13)(p - 1) \\[3ex] Substitute\:\: back \\[3ex] (12\cos\alpha + 13)(\cos\alpha - 1)$

(21.) $\dfrac{1}{\sin\omega - 1} - \dfrac{1}{\sin\omega + 1}$

$\dfrac{1}{\sin\omega - 1} - \dfrac{1}{\sin\omega + 1} \\[5ex] = \dfrac{1(\sin\omega + 1) - 1(\sin\omega - 1)}{(\sin\omega - 1)(\sin\omega + 1)} \\[5ex] (\sin\omega - 1)(\sin\omega + 1) = \sin^2\omega - 1^2 ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] = \dfrac{\sin\omega + 1 - \sin\omega + 1}{\sin^2\omega - 1^2} \\[5ex] = \dfrac{2}{\sin^2\omega - 1} \\[5ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \sin^2\omega - 1 = -\cos^2\omega \\[3ex] = \dfrac{2}{-\cos^2\omega} \\[5ex] = \dfrac{2}{-1 * \cos^2\omega} \\[5ex] \dfrac{1}{\cos\omega} = \sec\omega \\[5ex] \rightarrow \dfrac{1^2}{\cos^2\omega} = \dfrac{1}{\cos^2\omega} = \sec^2\omega \\[5ex] = 2 * -1 * \sec^2\omega \\[3ex] = -2\sec^2\omega$
(22.) Rationalize the numerator

$\sqrt{\dfrac{\sin \theta}{\cos^7 \theta}}$

We want to rationalize the numerator
This means that we want to make the numerator a rational number
What do we have to multiply by $1$ (the exponent at the numerator is $1$) in terms of exponent so that it is divisible by $2$
Why $2$? ... it is because of the square root.
Remember: Square root means an exponent of $\dfrac{1}{2}$... division by $2$
$\sqrt{x} = x^\frac{1}{2}$
So, we have to multiply $\sin \theta$ by $\sin \theta$ so we can get $\sin^2 \theta$ because $2$ is divisible by $2$
Whatever you do to the numerator, you have to do the same thing to the denominator
This is necessary in order to get an equivalent fraction to the original fraction
$\sqrt{\dfrac{\sin \theta}{\cos^7 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta}}{\sqrt{\cos^7 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta}}{\sqrt{\cos^7 \theta}} * \dfrac{\sqrt{\sin \theta}}{\sqrt{\sin \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta * \sin \theta}}{\sqrt{\cos^7 \theta * \sin \theta}} \\[5ex] = \dfrac{\sqrt{\sin^2 \theta}}{\sqrt{\cos^6 \theta * \cos \theta * \sin \theta}} \\[5ex] = \dfrac{\sin \theta}{\cos^3 \theta \sqrt{\cos \theta \sin \theta}} \\[5ex]$ The numerator is now a rational number.
(23.) Rationalize the denominator

$\sqrt{\dfrac{7 + 7\cos \theta}{7 - 7\cos \theta}}$

We want to rationalize the denominator
This means that we want to make the denominator a rational number
What do we have to multiply by $1$ (the exponent at the denominator is $1$) in terms of exponent so that it is divisible by $2$
Why $2$? ... it is because of the square root.
Remember: Square root means an exponent of $\dfrac{1}{2}$... division by $2$
$\sqrt{x} = x^\frac{1}{2}$
So, we have to multiply $7 - 7\cos \theta$ by $7 - 7\cos \theta$ so we can get the squared of it $(7 - 7\cos \theta)^2$ which is divisible by $2$
Whatever you do to the denominator, you have to do the same thing to the numerator
This is necessary in order to get an equivalent fraction to the original fraction
$\sqrt{\dfrac{7 + 7\cos \theta}{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} * \dfrac{\sqrt{7 - 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{(7 + 7\cos \theta) * (7 - 7\cos \theta)}}{\sqrt{(7 - 7\cos \theta)* (7 - 7\cos \theta)}} \\[5ex] (7 + 7\cos \theta) * (7 - 7\cos \theta) = 7^2 - (7\cos \theta)^2 ...\:Difference\: of\: Two\: Squares \\[3ex] (7 + 7\cos \theta) * (7 - 7\cos \theta) = 49 - 49\cos^2 \theta \\[3ex] = \dfrac{\sqrt{(49 - 49\cos^2 \theta)}}{\sqrt{(7 - 7\cos \theta)^2}} \\[5ex] = \dfrac{\sqrt{(49(1 - \cos^2 \theta)}}{(7 - 7\cos \theta)} \\[5ex] 1 - \cos^2 \theta = \sin^2 \theta ...\:Pythagorean\: Identity \\[3ex] = \dfrac{\sqrt{49\sin^2 \theta}}{7(1 - \cos \theta)} \\[5ex] = \dfrac{\sqrt{49} * \sqrt{\sin^2 \theta}}{7(1 - \cos \theta)} \\[5ex] = \dfrac{7\sin \theta}{7(1 - \cos \theta)} \\[5ex] = \dfrac{\sin \theta}{1 - \cos \theta} \\[5ex]$ The denominator is now a rational number.
(24.) $\dfrac{3 \sec \theta}{\csc \theta} + \dfrac{4\sin \theta}{\cos \theta}$

$\dfrac{3 \sec \theta}{\csc \theta} + \dfrac{4\sin \theta}{\cos \theta} \\[5ex] = \dfrac{3\sec \theta \cos \theta + 4\sin \theta \csc \theta}{\csc \theta \cos \theta} \\[5ex] \sec \theta = \dfrac{1}{\cos \theta} ... \: Reciprocal\: Identity \\[5ex] \csc \theta = \dfrac{1}{\sin \theta} ... \: Reciprocal\: Identity \\[5ex] = \dfrac{3 * \dfrac{1}{\cos \theta} * \cos \theta + 4 * \sin \theta * \dfrac{1}{\sin \theta}}{\dfrac{1}{\sin \theta} \cos \theta} \\[7ex] \tan \theta = \dfrac{\sin \theta}{\cos \theta} ... \: Quotient\: Identity \\[5ex] \cot \theta = \dfrac{\cos \theta}{\sin \theta} ... \: Quotient\: Identity \\[5ex] = \dfrac{7}{\dfrac{\cos \theta}{\sin \theta}} \\[7ex] = \dfrac{7}{\cot \theta} \\[5ex] \tan \theta = \dfrac{1}{\cot \theta} ... \:Reciprocal\: Identity \\[5ex] = 7\tan \theta$
(25.) $\cos3\omega + \cos6\omega$

$\cos3\omega + \cos6\omega \\[3ex] = 2\cos\left(\dfrac{3\omega + 6\omega}{2}\right)\cos\left(\dfrac{3\omega - 6\omega}{2}\right) ...Sum-to-Product\:\: Formula \\[5ex] = 2\cos\left(\dfrac{9\omega}{2}\right)\cos\left(\dfrac{-3\omega}{2}\right) \\[5ex] \cos(-\omega) = \cos(\omega) ... Even\:\: Identity \\[3ex] = 2\cos\left(\dfrac{9\omega}{2}\right)\cos\left(\dfrac{3\omega}{2}\right)$
(26.) $\cos(-\gamma) - \cos\gamma$

$\cos(-\gamma) - \cos\gamma \\[3ex] \cos(-\gamma) = \cos\gamma ... Even\:\: Identity \\[3ex] = \cos\gamma - \cos\gamma \\[3ex] = 0$
(27.) $\cos(\alpha + \beta) + \cos(\alpha - \beta)$

$\cos(\alpha + \beta) + \cos(\alpha - \beta) \\[3ex] \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta ... Sum\:\: Formula \\[3ex] \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta ... Difference\:\: Formula \\[3ex] = \cos\alpha\cos\beta - \sin\alpha\sin\beta + \cos\alpha\cos\beta + \sin\alpha\sin\beta \\[3ex] = 2\cos\alpha\cos\beta$
(28.) $\dfrac{\sin^5\theta\cos\theta}{\cos^5\theta\sin\theta}$

$\dfrac{\sin^5\theta\cos\theta}{\cos^5\theta\sin\theta} \\[5ex] = \dfrac{\sin^4\theta}{\cos^4\theta} \\[5ex] = \tan^4\theta$
(29.) $(3\cos\beta - 7\sin\beta)^2 + (7\cos\beta + 3\sin\beta)^2$

$(3\cos\beta - 7\sin\beta)^2 + (7\cos\beta + 3\sin\beta)^2 \\[3ex] \underline{augend} \\[3ex] (3\cos\beta - 7\sin\beta)^2 = (3\cos\beta - 7\sin\beta)(3\cos\beta - 7\sin\beta) \\[3ex] = 9\cos^2\beta - 21\sin\beta\cos\beta - 21\sin\beta\cos\beta + 49\sin^2\beta \\[3ex] = 9\cos^2\beta - 42\sin\beta\cos\beta + 49\sin^2\beta \\[3ex] \underline{added} \\[3ex] (7\cos\beta + 3\sin\beta)^2 = (7\cos\beta + 3\sin\beta)(7\cos\beta + 3\sin\beta) \\[3ex] = 49\cos^2\beta + 21\sin\beta\cos\beta + 21\sin\beta\cos\beta + 9\sin^2\beta \\[3ex] = 49\cos^2\beta + 42\sin\beta\cos\beta + 9\sin^2\beta \\[3ex] \underline{sum} \\[3ex] = 9\cos^2\beta - 42\sin\beta\cos\beta + 49\sin^2\beta + 49\cos^2\beta + 42\sin\beta\cos\beta + 9\sin^2\beta \\[3ex] = 58\cos^2\beta + 58\sin^2\beta \\[3ex] = 58\sin^2\beta + 58\cos^2\beta \\[3ex] = 58(\sin^2\beta + \cos^2\beta) \\[3ex] \sin^2\beta + \cos^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] = 58 * 1 \\[3ex] = 58$
(30.) $\dfrac{3\sec\gamma}{\csc\gamma} + \dfrac{4\sin\gamma}{\cos\gamma}$

$\dfrac{3\sec\gamma}{\csc\gamma} + \dfrac{4\sin\gamma}{\cos\gamma} \\[5ex] = \dfrac{3\sec\gamma\cos\gamma}{\csc\gamma\cos\gamma} + \dfrac{4\sin\gamma\csc\gamma}{\csc\gamma\cos\gamma} \\[5ex] = \dfrac{3\sec\gamma\cos\gamma + 4\sin\gamma\csc\gamma}{\csc\gamma\cos\gamma} \\[5ex] \underline{Numerator} \\[3ex] \sec\gamma = \dfrac{1}{\cos\gamma} ... Reciprocal\:\: Identity \\[5ex] 3\sec\gamma\cos\gamma = 3 * \dfrac{1}{\cos\gamma} * \cos\gamma = 3 * 1 = 3 \\[5ex] \csc\gamma = \dfrac{1}{\sin\gamma} ... Reciprocal\:\: Identity \\[5ex] 4\sin\gamma\csc\gamma = 4 * \sin\gamma * \dfrac{1}{\sin\gamma} = 4 * 1 = 4 \\[5ex] 3 + 4 = 7 \\[3ex] \underline{Denominator} \\[3ex] \csc\gamma\cos\gamma = \dfrac{1}{\sin\gamma} * \cos\gamma = \dfrac{\cos\gamma}{\sin\gamma} \\[5ex] \dfrac{\cos\gamma}{\sin\gamma} = \cot\gamma ... Quotient\:\: Identity \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{7}{\cot\gamma} \\[5ex] = 7 * \dfrac{1}{\cot\gamma} \\[5ex] \dfrac{1}{\cot\gamma} = \tan\gamma ... Reciprocal\:\: Identity \\[5ex] = 7 * \tan\gamma \\[3ex] = 7\tan\gamma$
(31.) $\dfrac{5\sin\alpha}{\cos^2\alpha} * \dfrac{\cos^2\alpha - \cos\alpha\sin\alpha}{\cos^2\alpha - \sin^2\alpha}$

$\dfrac{5\sin\alpha}{\cos^2\alpha} * \dfrac{\cos^2\alpha - \cos\alpha\sin\alpha}{\cos^2\alpha - \sin^2\alpha} \\[5ex] \cos^2\alpha - \cos\alpha\sin\alpha = \cos\alpha(\cos\alpha - \sin\alpha) ... Factor\:\: by\:\: GCF \\[3ex] \cos^2\alpha - \sin^2\alpha = (\cos\alpha + \sin\alpha)(\cos\alpha - \sin\alpha) ... Difference \:\:of\:\: Two\:\: Squares \\[3ex] = \dfrac{5\sin\alpha}{\cos^2\alpha} * \dfrac{\cos\alpha(\cos\alpha - \sin\alpha)}{(\cos\alpha + \sin\alpha)(\cos\alpha - \sin\alpha)} \\[5ex] = \dfrac{5\sin\alpha}{\cos\alpha} * \dfrac{1}{\cos\alpha + \sin\alpha} \\[5ex] \dfrac{\sin\alpha}{\cos\alpha} = \tan\alpha ... Quotient\:\: Identity \\[5ex] = 5\tan\alpha * \dfrac{1}{\cos\alpha + \sin\alpha} \\[5ex] = \dfrac{5\tan\alpha * 1}{\cos\alpha + \sin\alpha} \\[5ex] = \dfrac{5\tan\alpha}{\cos\alpha + \sin\alpha}$
(32.) $\sqrt{\sin^4\theta\cos^2\theta} * \sqrt{\cos^2\theta}$

$\sqrt{\sin^4\theta\cos^2\theta} * \sqrt{\cos^2\theta} \\[5ex] \underline{multiplier} \\[3ex] \sqrt{\sin^4\theta\cos^2\theta} = \left(\sin^4\theta\right)^\dfrac{1}{2} * \left(\cos^2\theta\right)^\dfrac{1}{2} \\[5ex] = \sin^2\theta * \cos\theta \\[3ex] \underline{multiplicand} \\[3ex] \sqrt{\cos^2\theta} = \left(\cos^2\theta\right)^\dfrac{1}{2} \\[5ex] = \cos\theta \\[3ex] \underline{product} \\[3ex] = \sin^2\theta * \cos\theta * \cos\theta \\[3ex] = \sin^2\theta\cos^2\theta$
(33.) For $\sqrt{p^2 - c^2}$, determine $\sin\beta$ if $c = p\cos\beta$

$c = p\cos\beta \\[3ex] \sqrt{p^2 - c^2} = \sqrt{p^2 - (p\cos\beta)^2} \\[3ex] = \sqrt{p^2 - p^2\cos^2\beta} \\[3ex] = \sqrt{p^2(1 - \cos^2\beta)} \\[3ex] \sin^2\beta + \cos^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] \sin^2\beta = 1 - \cos^2\beta \\[3ex] = \sqrt{p^2 * \sin^2\beta} \\[3ex] = p\sin\beta \\[3ex] \therefore \sqrt{p^2 - c^2} = p\sin\beta \\[3ex] p\sin\beta = \sqrt{p^2 - c^2} \\[3ex] \sin\beta = \dfrac{\sqrt{p^2 - c^2}}{p}$
(34.) $\sqrt{\dfrac{\sin^2\phi}{2\cos^2\phi}}$

$\sqrt{\dfrac{\sin^2\phi}{2\cos^2\phi}} \\[5ex] = \dfrac{\sqrt{\sin^2\phi}}{\sqrt{2\cos^2\phi}} \\[5ex] = \dfrac{\sin\phi}{\sqrt{2}\cos\phi} \\[5ex] = \dfrac{\sin\phi}{\cos\phi} * \dfrac{1}{\sqrt{2}} \\[5ex] \dfrac{\sin\phi}{\cos\phi} = \tan\phi ... Quotient\:\: Identity \\[5ex] = \tan\phi * \dfrac{1}{\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \\[5ex] = \dfrac{\sqrt{2}\tan\phi}{2}$
(35.) For $\sqrt{p^2 - c^2}$, determine $\tan\beta$ if $c = p\cos\beta$

$c = p\cos\beta ...eqn.(1) \\[3ex] \sqrt{p^2 - c^2} = \sqrt{p^2 - (p\cos\beta)^2} \\[3ex] = \sqrt{p^2 - p^2\cos^2\beta} \\[3ex] = \sqrt{p^2(1 - \cos^2\beta)} \\[3ex] \sin^2\beta + \cos^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] \sin^2\beta = 1 - \cos^2\beta \\[3ex] = \sqrt{p^2 * \sin^2\beta} \\[3ex] = p\sin\beta \\[3ex] \therefore \sqrt{p^2 - c^2} = p\sin\beta \\[3ex] p\sin\beta = \sqrt{p^2 - c^2} \\[3ex] \sin\beta = \dfrac{\sqrt{p^2 - c^2}}{p} ...eqn.(2) \\[5ex] \tan\beta = \dfrac{\sin\beta}{\cos\beta} ... Quotient\:\: Identity \\[5ex] Divide\:\: both\:\: sides\:\: of\:\: eqn.(2) \:\:by\:\: \cos\beta \\[3ex] \dfrac{\sin\beta}{\cos\beta} = \dfrac{\sqrt{p^2 - c^2}}{p} \div \cos\beta \\[5ex] \tan\beta = \dfrac{\sqrt{p^2 - c^2}}{p} \div \cos\beta ...eqn.(3) \\[5ex] From\:\: eqn.(1) \\[3ex] p\cos\beta = c \\[3ex] \cos\beta = \dfrac{c}{p} \\[5ex] Substitute\:\: for\:\: \cos\beta \:\:in\:\: eqn. (3) \\[3ex] \tan\beta = \dfrac{\sqrt{p^2 - c^2}}{p} \div \dfrac{c}{p} \\[5ex] \tan\beta = \dfrac{\sqrt{p^2 - c^2}}{p} * \dfrac{p}{c} \\[5ex] \tan\beta = \dfrac{\sqrt{p^2 - c^2}}{c}$
(36.) $\dfrac{\sin\gamma + \cos\gamma}{\sin\gamma} + \dfrac{\sin\gamma - \cos\gamma}{\cos\gamma}$

$\dfrac{\sin\gamma + \cos\gamma}{\sin\gamma} + \dfrac{\sin\gamma - \cos\gamma}{\cos\gamma} \\[5ex] = \dfrac{\cos\gamma(\sin\gamma + \cos\gamma) + \sin\gamma(\sin\gamma - \cos\gamma)}{\sin\gamma\cos\gamma} \\[5ex] = \dfrac{\cos\gamma\sin\gamma + \cos^2\gamma + \sin^\gamma - \sin\gamma\cos\gamma)}{\sin\gamma\cos\gamma} \\[5ex] \sin^2\gamma + \cos^2\gamma = 1 ... Pythagorean\:\: Identity \\[3ex] = \dfrac{1}{\sin\gamma\cos\gamma} \\[5ex] = \dfrac{1}{\sin\gamma} * \dfrac{1}{\cos\gamma} \\[5ex] \dfrac{1}{\sin\gamma} = \csc\gamma ... Reciprocal\:\: Identity \\[5ex] \dfrac{1}{\cos\gamma} = \sec\gamma ... Reciprocal\:\: Identity \\[5ex] = \csc\gamma\sec\gamma$
(37.) $\dfrac{1 + \tan\beta}{1 - \tan\beta} + \dfrac{1 + \cot\beta}{1 - \cot\beta}$

$\dfrac{1 + \tan\beta}{1 - \tan\beta} + \dfrac{1 + \cot\beta}{1 - \cot\beta} \\[5ex] \underline{First\:\: term} \\[3ex] \dfrac{1 + \tan\beta}{1 - \tan\beta} \\[5ex] \tan\beta = \dfrac{\sin\beta}{\cos\beta} ...Quotient\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] 1 + \tan\beta = 1 + \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta}{\cos\beta} + \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta} \\[5ex] \underline{Denominator} \\[3ex] 1 - \tan\beta = 1 - \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta}{\cos\beta} - \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\cos\beta} \\[5ex] \dfrac{Numerator}{Denominator} \\[3ex] = (1 + \tan\beta) \div (1 - \tan\beta) \\[3ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta} \div \dfrac{\cos\beta - \sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta} * \dfrac{\cos\beta}{\cos\beta - \sin\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta - \sin\beta} \\[5ex] \underline{Second\:\: term} \\[3ex] \dfrac{1 + \cot\beta}{1 - \cot\beta} \\[5ex] \cot\beta = \dfrac{\cos\beta}{\sin\beta} ...Quotient\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] 1 + \cot\beta = 1 + \dfrac{\cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta}{\sin\beta} + \dfrac{\cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta + \cos\beta}{\sin\beta} \\[5ex] \underline{Denominator} \\[3ex] 1 - \cot\beta = 1 - \dfrac{\cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta}{\sin\beta} - \dfrac{\cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta - \cos\beta}{\sin\beta} \\[5ex] \dfrac{Numerator}{Denominator} \\[3ex] = (1 + \cot\beta) \div (1 - \cot\beta) \\[3ex] = \dfrac{\sin\beta + \cos\beta}{\sin\beta} \div \dfrac{\sin\beta - \cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta + \cos\beta}{\sin\beta} * \dfrac{\sin\beta}{\sin\beta - \cos\beta} \\[5ex] = \dfrac{\sin\beta + \cos\beta}{\sin\beta - \cos\beta} \\[5ex] \underline{First\:\: term\:\: + \:\:Second\:\: term} \\[3ex] \dfrac{1 + \tan\beta}{1 - \tan\beta} + \dfrac{1 + \cot\beta}{1 - \cot\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta - \sin\beta} + \dfrac{\sin\beta + \cos\beta}{\sin\beta - \cos\beta} \\[5ex] = \dfrac{(\cos\beta + \sin\beta)(\sin\beta - \cos\beta) + (\sin\beta + \cos\beta)(\cos\beta - \sin\beta)}{(\cos\beta - \sin\beta)(\sin\beta - \cos\beta)} \\[5ex] = \dfrac{\sin\beta\cos\beta - \cos^2\beta + \sin^2\beta - \sin\beta\cos\beta + \sin\beta\cos\beta - \sin^2\beta + \cos^2\beta - \sin\beta\cos\beta}{(\cos\beta - \sin\beta)(\sin\beta - \cos\beta)} \\[5ex] = \dfrac{0}{(\cos\beta - \sin\beta)(\sin\beta - \cos\beta)} \\[5ex] = 0$
(38.) $(1 + \cot^2\beta)\sin^2\beta$

$(1 + \cot^2\beta)\sin^2\beta \\[3ex] = \sin^2\beta(1 + \cot^2\beta) \\[3ex] \cot\beta = \dfrac{\cos\beta}{\sin\beta} ... Quotient\:\: Identity \\[5ex] \cot^2\beta = \left(\dfrac{\cos\beta}{\sin\beta}\right)^2 = \dfrac{\cos^2\beta}{\sin^2\beta} \\[5ex] = \sin^2\beta\left(1 + \dfrac{\cos^2\beta}{\sin^2\beta}\right) \\[5ex] = \sin^2\beta(1) + \sin^2\beta\left(\dfrac{\cos^2\beta}{\sin^2\beta}\right) \\[5ex] = \sin^2\beta + \cos^2\beta \\[3ex] = 1 ... Pythagorean\:\: Identity$
(39.) $\cos(\alpha - \beta)\cos\beta - \sin(\alpha - \beta)\sin\beta$

$\cos(\alpha - \beta)\cos\beta - \sin(\alpha - \beta)\sin\beta \\[3ex] = \cos\beta * \cos(\alpha - \beta) - \sin\beta * \sin(\alpha - \beta) \\[3ex] \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta ... Difference\:\: Formula \\[3ex] \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta ... Difference\:\: Formula \\[3ex] = \cos\beta(\cos\alpha\cos\beta + \sin\alpha\sin\beta) - \sin\beta(\sin\alpha\cos\beta - \cos\alpha\sin\beta) \\[3ex] = \cos\alpha\cos^2\beta + \sin\alpha\sin\beta\cos\beta - \sin\alpha\sin\beta\cos\beta + \cos\alpha\sin^2\beta \\[3ex] = \cos\alpha\cos^2\beta + \cos\alpha\sin^2\beta \\[3ex] = \cos\alpha(\cos^\beta + \sin^2\beta) \\[3ex] \cos^2\beta + \sin^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] = \cos\alpha * 1 \\[3ex] = \cos\alpha$
(40.) $\cot\alpha * \sec\alpha$

$\cot\alpha * \sec\alpha \\[3ex] \cot\alpha = \dfrac{\cos\alpha}{\sin\alpha} ... Quotient\:\: Identity \\[5ex] \sec\alpha = \dfrac{1}{\cos\alpha} ... Reciprocal\:\: Identity \\[5ex] = \dfrac{\cos\alpha}{\sin\alpha} * \dfrac{1}{\cos\alpha} \\[5ex] = \dfrac{1}{\sin\alpha} \\[5ex] = \csc\alpha$

(41.) Determine the exact value of
$\sin1^\circ * \sin2^\circ * \sin3^\circ * ... * \sin44^\circ * \sin45^\circ * \sec46^\circ * ... * \sec87^\circ * \sec88^\circ * \sec89^\circ$

Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the sine function is the cosecant function
(3.) The cosecant function and the secant function are cofunctions
(4.) $\csc1$ and $\sec89$ are cofunctions; $\csc2$ and $\sec88$ are cofunctions; $\csc3$ and $\sec87$ are cofunctions, $\csc44$ and $\sec46$ are cofunctions, etc.
(5.) It starts from $\sin1$ up to $\sin45$, then it changes to $\sec46$ and continues till $\sec89$
(6.) Only the reciprocal of $\sin45$ does not have an included cofunction
(7.) Every other term besides $\sin45$ will cancel out

$\sin1^\circ * \sin2^\circ * \sin3^\circ * ... * \sin44^\circ * \sin45^\circ * \sec46^\circ * ... * \sec87^\circ * \sec88^\circ * \sec89^\circ \\[3ex] \dfrac{1}{\sin1} = \csc1 ... Reciprocal\:\: Identity \\[5ex] \sec89 = \csc1 ... Cofunction\:\: Identity \\[3ex] \therefore \sec89 = \dfrac{1}{\sin1} ... Equality \:\:of\:\: \csc1 \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \sin45 \\[3ex] = \sin1 * \sin2 * \sin3 * ... * \sin44 * \sin45 * \dfrac{1}{\sin44} * ... * \dfrac{1}{\sin3} * \dfrac{1}{\sin2} * \dfrac{1}{\sin1} \\[5ex] = \sin45 \\[3ex] = \dfrac{\sqrt{2}}{2} ... Unit\:\: Circle\:\: Trigonometry$
(42.) Determine the exact value of
$\cos1^\circ * \cos2^\circ * \cos3^\circ * ... * \cos44^\circ * \cos45^\circ * \csc46^\circ * ... * \csc87^\circ * \csc88^\circ * \csc89^\circ$

Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the cosine function is the secant function
(3.) The secant function and the cosecant function are cofunctions
(4.) $\sec1$ and $\csc89$ are cofunctions; $\sec2$ and $\csc88$ are cofunctions; $\sec3$ and $\csc87$ are cofunctions, $\sec44$ and $\csc46$ are cofunctions, etc.
(5.) It starts from $\cos1$ up to $\cos45$, then it changes to $\csc46$ and continues till $\csc89$
(6.) Only the reciprocal of $\cos45$ does not have an included cofunction
(7.) Every other term besides $\cos45$ will cancel out

$\cos1^\circ * \cos2^\circ * \cos3^\circ * ... * \cos44^\circ * \cos45^\circ * \csc46^\circ * ... * \csc87^\circ * \csc88^\circ * \csc89^\circ \\[3ex] \dfrac{1}{\cos1} = \sec1 ... Reciprocal\:\: Identity \\[5ex] \csc89 = \sec1 ... Cofunction\:\: Identity \\[3ex] \therefore \csc89 = \dfrac{1}{\cos1} ... Equality \:\:of\:\: \sec1 \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \cos45 \\[3ex] = \cos1 * \cos2 * \cos3 * ... * \cos44 * \cos45 * \dfrac{1}{\cos44} * ... * \dfrac{1}{\cos3} * \dfrac{1}{\cos2} * \dfrac{1}{\cos1} \\[5ex] = \cos45 \\[3ex] = \dfrac{\sqrt{2}}{2} ... Unit\:\: Circle\:\: Trigonometry$
(43.) Determine the exact value of
$\tan1^\circ * \tan2^\circ * \tan3^\circ * ... * \tan87^\circ * \tan88^\circ * \tan89^\circ$

Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the tangent function is the cotangent function
(3.) The tangent function and the cotangent function are cofunctions
(4.) $\tan89$ and $\cot1$ are cofunctions; $\tan88$ and $\cot2$ are cofunctions; $\tan87$ and $\cot3$ are cofunctions etc.
(5.) It starts from $\tan1$ up to $\tan89$
(6.) Only the reciprocal of $\tan45$ does not have an included cofunction
(7.) Every term besides $\tan45$ will cancel out

$\tan1^\circ * \tan2^\circ * \tan3^\circ * ... * \tan87^\circ * \tan88^\circ * \tan89^\circ \\[3ex] \dfrac{1}{\tan1} = \cot1 ... Reciprocal\:\: Identity \\[5ex] \tan89 = \cot1 ... Cofunction\:\: Identity \\[3ex] \therefore \tan89 = \dfrac{1}{\tan1} ... Equality \:\:of\:\: \cot1 \\[5ex] Similarly\:\: \tan88 = \dfrac{1}{\tan2} \\[5ex] Similarly\:\: \tan87 = \dfrac{1}{\tan3} \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \tan45 \\[3ex] = \tan1 * \tan2 * \tan3 * ... * \tan45 * \dfrac{1}{\tan3} * \dfrac{1}{\tan2} * \dfrac{1}{\tan1} \\[5ex] = \tan45 \\[3ex] = 1 ... Unit\:\: Circle\:\: Trigonometry$
(44.) Determine the exact value of
$\cot1^\circ * \cot2^\circ * \cot3^\circ * ... * \cot87^\circ * \cot88^\circ * \cot89^\circ$

Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the cotangent function is the tangent function
(3.) The cotangent function and the tangent function are cofunctions
(4.) $\cot89$ and $\tan1$ are cofunctions; $\cot88$ and $\tan2$ are cofunctions; $\cot87$ and $\tan3$ are cofunctions etc.
(5.) It starts from $\cot1$ up to $\cot89$
(6.) Only the reciprocal of $\cot45$ does not have an included cofunction
(7.) Every term besides $\cot45$ will cancel out

$\cot1^\circ * \cot2^\circ * \cot3^\circ * ... * \cot87^\circ * \cot88^\circ * \cot89^\circ \\[3ex] \dfrac{1}{\cot1} = \tan1 ... From\:\: Reciprocal\:\: Identity \\[5ex] \cot89 = \tan1 ... Cofunction\:\: Identity \\[3ex] \therefore \cot89 = \dfrac{1}{\cot1} ... Equality \:\:of\:\: \tan1 \\[5ex] Similarly\:\: \cot88 = \dfrac{1}{\cot2} \\[5ex] Similarly\:\: \cot87 = \dfrac{1}{\cot3} \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \cot45 \\[3ex] = \cot1 * \cot2 * \cot3 * ... * \cot45 * \dfrac{1}{\cot3} * \dfrac{1}{\cot2} * \dfrac{1}{\cot1} \\[5ex] = \cot45 \\[3ex] = \dfrac{1}{\tan45} \\[5ex] \tan45 = 1 ... Unit\:\: Circle\:\: Trigonometry \\[3ex] = \dfrac{1}{1} \\[5ex] = 1$
(45.) Determine the exact value of
$\csc1^\circ * \csc2^\circ * \csc3^\circ * ... * \csc44^\circ * \csc45^\circ * \cos46^\circ * ... * \cos87^\circ * \cos88^\circ * \cos89^\circ$

Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the cosecant function is the sine function
(3.) The sine function and the cosine function are cofunctions
(4.) $\sin1$ and $\cos89$ are cofunctions; $\sin2$ and $\cos88$ are cofunctions; $\sin3$ and $\cos87$ are cofunctions, $\sin44$ and $\cos46$ are cofunctions, etc.
(5.) It starts from $\csc1$ up to $\csc45$, then it changes to $\cos46$ and continues till $\cos89$
(6.) Only the reciprocal of $\csc45$ does not have an included cofunction
(7.) Every other term besides $\csc45$ will cancel out

$\csc1^\circ * \csc2^\circ * \csc3^\circ * ... * \csc44^\circ * \csc45^\circ * \cos46^\circ * ... * \cos87^\circ * \cos88^\circ * \cos89^\circ \\[3ex] \dfrac{1}{\csc1} = \sin1 ... From\:\: Reciprocal\:\: Identity \\[5ex] \cos89 = \sin1 ... Cofunction\:\: Identity \\[3ex] \therefore \cos89 = \dfrac{1}{\csc1} ... Equality \:\:of\:\: \sin1 \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \csc45 \\[3ex] = \csc1 * \csc2 * \csc3 * ... * \csc44 * \csc45 * \dfrac{1}{\csc44} * ... * \dfrac{1}{\csc3} * \dfrac{1}{\csc2} * \dfrac{1}{\csc1} \\[5ex] = \csc45 \\[3ex] = \dfrac{1}{\sin45} ... Reciprocal\:\: Identity \\[5ex] \sin45 = \dfrac{\sqrt{2}}{2} ... Unit\:\: Circle\:\: Trigonometry \\[5ex] = 1 \div \dfrac{\sqrt{2}}{2} \\[5ex] = 1 * \dfrac{2}{\sqrt{2}} \\[5ex] = \dfrac{2}{\sqrt{2}}$
(46.) Determine the exact value of
$\sec1^\circ * \sec2^\circ * \sec3^\circ * ... * \sec44^\circ * \sec45^\circ * \sin46^\circ * ... * \sin87^\circ * \sin88^\circ * \sin89^\circ$

Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the secant function is the cosine function
(3.) The cosine function and the sine function are cofunctions
(4.) $\cos1$ and $\sin89$ are cofunctions; $\cos2$ and $\sin88$ are cofunctions; $\cos3$ and $\sin87$ are cofunctions, $\cos44$ and $\sin46$ are cofunctions, etc.
(5.) It starts from $\sec1$ up to $\sec45$, then it changes to $\sin46$ and continues till $\sin89$
(6.) Only the reciprocal of $\sec45$ does not have an included cofunction
(7.) Every other term besides $\sec45$ will cancel out

$\sec1^\circ * \sec2^\circ * \sec3^\circ * ... * \sec44^\circ * \sec45^\circ * \sin46^\circ * ... * \sin87^\circ * \sin88^\circ * \sin89^\circ \\[3ex] \dfrac{1}{\sec1} = \cos1 ... Reciprocal\:\: Identity \\[5ex] \sin89 = \cos1 ... Cofunction\:\: Identity \\[3ex] \therefore \sin89 = \dfrac{1}{\sec1} ... Equality \:\:of\:\: \cos1 \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \sec45 \\[3ex] = \sec1 * \sec2 * \sec3 * ... * \sec44 * \sec45 * \dfrac{1}{\sec44} * ... * \dfrac{1}{\sec3} * \dfrac{1}{\sec2} * \dfrac{1}{\sec1} \\[5ex] = \sec45 \\[3ex] = \dfrac{1}{\cos45} ... Reciprocal\:\: Identity \\[5ex] \cos45 = \dfrac{\sqrt{2}}{2} ... Unit\:\: Circle\:\: Trigonometry \\[5ex] = 1 \div \dfrac{\sqrt{2}}{2} \\[5ex] = 1 * \dfrac{2}{\sqrt{2}} \\[5ex] = \dfrac{2}{\sqrt{2}}$
(47.) $\dfrac{\sin^3\omega + \cos^3\omega}{\sin\omega + \cos\omega}$

$\dfrac{\sin^3\omega + \cos^3\omega}{\sin\omega + \cos\omega} \\[5ex] \sin^3\omega + \cos^3\omega = (\sin\omega + \cos\omega)(\sin^2\omega - \sin\omega\cos\omega + \cos^2\omega) ... Sum\:\: of\:\: Two\:\: Cubes \\[3ex] = \dfrac{(\sin\omega + \cos\omega)(\sin^2\omega - \sin\omega\cos\omega + \cos^2\omega)}{\sin\omega + \cos\omega} \\[5ex] = \sin^2\omega - \sin\omega\cos\omega + \cos^2\omega \\[3ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] = 1 - \sin\omega\cos\omega$
(48.) $\dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega}$

$\dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] \sec\omega = \dfrac{1}{\cos\omega} ... Reciprocal\:\: Identity \\[5ex] \tan\omega = \dfrac{\sin\omega}{\cos\omega} ... Quotient\:\: Identity \\[5ex] \sec\omega + \tan\omega = \dfrac{1}{\cos\omega} + \dfrac{\sin\omega}{\cos\omega} = \dfrac{1 + \sin\omega}{\cos\omega} \\[5ex] \underline{Minuend} \\[3ex] \dfrac{\sec\omega + \tan\omega}{\tan\omega} \\[5ex] = (\sec\omega + \tan\omega) \div \tan\omega \\[3ex] = \dfrac{1 + \sin\omega}{\cos\omega} \div \dfrac{\sin\omega}{\cos\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\cos\omega} * \dfrac{\cos\omega}{\sin\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\sin\omega} \\[5ex] \underline{Subtrahend} \\[3ex] \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] = (\sec\omega + \tan\omega) \div \sec\omega \\[3ex] = \dfrac{1 + \sin\omega}{\cos\omega} \div \dfrac{1}{\cos\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\cos\omega} * \dfrac{\cos\omega}{1} \\[5ex] = 1 + \sin\omega \\[3ex] \rightarrow \dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\sin\omega} - (1 + \sin\omega) \\[5ex] = \dfrac{1}{\sin\omega} + \dfrac{\sin\omega}{\sin\omega} - 1 - \sin\omega \\[5ex] = \dfrac{1}{\sin\omega} + 1 - 1 - \sin\omega \\[5ex] = \csc\omega - \sin\omega$
(49.) $\sin7\omega - \sin9\omega$

$\sin7\omega - \sin9\omega \\[3ex] = 2\sin\left(\dfrac{7\omega - 9\omega}{2}\right)\cos\left(\dfrac{7\omega + 9\omega}{2}\right) ...Sum-to-Product\:\: Formula \\[5ex] = 2\sin\left(\dfrac{-2\omega}{2}\right)\cos\left(\dfrac{16\omega}{2}\right) \\[5ex] = 2\sin(-\omega)\cos(8\omega) \\[3ex] \sin(-\omega) = -\sin(\omega) ... Odd\:\: Identity \\[3ex] = 2 * -\sin(\omega) * \cos(8\omega) \\[3ex] = -2\sin\omega\cos8\omega$
(50.) $\dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega}$

$\dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] \sec\omega = \dfrac{1}{\cos\omega} ... Reciprocal\:\: Identity \\[5ex] \tan\omega = \dfrac{\sin\omega}{\cos\omega} ... Quotient\:\: Identity \\[5ex] \sec\omega + \tan\omega = \dfrac{1}{\cos\omega} + \dfrac{\sin\omega}{\cos\omega} = \dfrac{1 + \sin\omega}{\cos\omega} \\[5ex] \underline{Minuend} \\[3ex] \dfrac{\sec\omega + \tan\omega}{\tan\omega} \\[5ex] = (\sec\omega + \tan\omega) \div \tan\omega \\[3ex] = \dfrac{1 + \sin\omega}{\cos\omega} \div \dfrac{\sin\omega}{\cos\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\cos\omega} * \dfrac{\cos\omega}{\sin\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\sin\omega} \\[5ex] \underline{Subtrahend} \\[3ex] \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] = (\sec\omega + \tan\omega) \div \sec\omega \\[3ex] = \dfrac{1 + \sin\omega}{\cos\omega} \div \dfrac{1}{\cos\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\cos\omega} * \dfrac{\cos\omega}{1} \\[5ex] = 1 + \sin\omega \\[3ex] \rightarrow \dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\sin\omega} - (1 + \sin\omega) \\[5ex] = \dfrac{1}{\sin\omega} + \dfrac{\sin\omega}{\sin\omega} - 1 - \sin\omega \\[5ex] = \dfrac{1}{\sin\omega} + 1 - 1 - \sin\omega \\[5ex] = \dfrac{1}{\sin\omega} - \dfrac{\sin\omega}{1} \\[5ex] = \dfrac{1 - \sin\omega(\sin\omega)}{\sin\omega} \\[5ex] = \dfrac{1 - \sin^2\omega}{\sin\omega} \\[5ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \cos^2\omega = 1 - \sin^2\omega \\[3ex] = \dfrac{\cos^2\omega}{\sin\omega} \\[5ex] = \dfrac{\cos\omega * \cos\omega}{\sin\omega} \\[5ex] = \cos\omega * \dfrac{\cos\omega}{\sin\omega} \\[5ex] \cot\omega = \dfrac{\cos\omega}{\sin\omega} ... Quotient\:\: Identity \\[5ex] = \cos\omega * \cot\omega \\[3ex] = \cos\omega\cot\omega$
(51.) $7\cos^4\theta - 7\sin^4\theta$

$7\cos^4\theta - 7\sin^4\theta \\[3ex] = 7(\cos^4\theta - \sin^4\theta) \\[3ex] \underline{Multiplicand} \\[3ex] \cos^4\theta - \sin^4\theta \\[3ex] = (\cos^2\theta)^2 - (\sin^2\theta)^2 \\[3ex] = (\cos^2\theta + \sin^2\theta)(\cos^2\theta - \sin^2\theta) ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] \cos^2\theta + \sin^2\theta = 1 ... Pythagorean\:\: Identity \\[3ex] = 1 * \cos^2\theta - \sin^2\theta \\[3ex] = \cos^2\theta - \sin^2\theta \\[3ex] = \cos2\theta ... Double-Angle\:\: Identity \\[3ex] \rightarrow 7 * \cos2\theta \\[3ex] = 7\cos2\theta$
(52.) $\tan^2\beta - \sec^2\beta$

$\tan^2\beta - \sec^2\beta \\[3ex] 1 + \tan^2\beta = \sec^2\beta ... Pythagorean\:\: Identity \\[3ex] \tan^2\beta = \sec^2\beta - 1 \\[3ex] = \sec^2\beta - 1 - \sec^2\beta \\[3ex] = \sec^2\beta - \sec^2\beta - 1 \\[3ex] = -1$
(53.) $\sin(\theta + \omega) - \sin(\theta - \omega)$

$\sin(\theta + \omega) - \sin(\theta - \omega) \\[3ex] \sin(\theta + \omega) = \sin\theta\cos\omega + \cos\theta\sin\omega ... Sum\:\: Formula \\[3ex] \sin(\theta - \omega) = \sin\theta\cos\omega - \cos\theta\sin\omega ... Difference\:\: Formula \\[3ex] = \sin\theta\cos\omega + \cos\theta\sin\omega - (\sin\theta\cos\omega - \cos\theta\sin\omega) \\[3ex] = \sin\theta\cos\omega + \cos\theta\sin\omega - \sin\theta\cos\omega + \cos\theta\sin\omega \\[3ex] = \cos\theta\sin\omega + \cos\theta\sin\omega \\[3ex] = 2\cos\theta\sin\omega$
(54.) $(1 + \cot \beta)^2$

$(1 + \cot \beta)^2 = (1 + \cot \beta)(1 + \cot \beta) \\[3ex] = 1^2 + \cot \beta + \cot \beta + \cot^2 \beta \\[3ex] = 1 + 2\cot \beta + \cot^2 \beta \\[3ex] = 1 + \cot^2 \beta + 2\cot \beta \\[3ex] 1 + \cot^2 \beta = \csc^2 \beta ... Pythagorean\:\: Identity \\[3ex] = \csc^2 \beta + 2\cot \beta$
(55.) $\dfrac{1 + \sin^2\omega}{1 - \sin^2\omega} + 1$

$\dfrac{1 + \sin^2\omega}{1 - \sin^2\omega} + 1 \\[5ex] = \dfrac{1 + \sin^2\omega}{1 - \sin^2\omega} + \dfrac{1 - \sin^2\omega}{1 - \sin^2\omega} \\[5ex] = \dfrac{(1 + \sin^2\omega) + (1 - \sin^2\omega)}{1 - \sin^2\omega} \\[5ex] = \dfrac{1 + \sin^2\omega + 1 - \sin^2\omega}{1 - \sin^2\omega} \\[5ex] = \dfrac{2}{1 - \sin^2\omega} \\[5ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \cos^2\omega = 1 - \sin^2\omega \\[3ex] = \dfrac{2}{\cos^2\omega} \\[5ex] \sec\omega = \dfrac{1}{\cos\omega} ... Reciprocal\:\: Identity \\[5ex] \rightarrow \sec^2\omega = \dfrac{1^2}{\cos^2\omega} = \dfrac{1}{\cos^2\omega} \\[5ex] = 2 * \dfrac{1}{\cos^2\omega} \\[5ex] = 2\sec^2\omega$
(56.) $\dfrac{\sec\alpha - \csc\alpha}{\sec\alpha\csc\alpha}$

$\dfrac{\sec\alpha - \csc\alpha}{\sec\alpha\csc\alpha} \\[5ex] \sec\alpha = \dfrac{1}{\cos\alpha} ... Reciprocal\:\: Identity \\[5ex] \csc\alpha = \dfrac{1}{\sin\alpha} ... Reciprocal\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] \sec\alpha - \csc\alpha \\[3ex] = \dfrac{1}{\cos\alpha} - \dfrac{1}{\sin\alpha} \\[5ex] = \dfrac{\sin\alpha - \cos\alpha}{\sin\alpha\cos\alpha} \\[5ex] \underline{Denominator} \\[3ex] \sec\alpha\csc\alpha \\[3ex] = \dfrac{1}{\cos\alpha} * \dfrac{1}{\sin\alpha} \\[5ex] = \dfrac{1}{\sin\alpha\cos\alpha} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{\sin\alpha - \cos\alpha}{\sin\alpha\cos\alpha} \div \dfrac{1}{\sin\alpha\cos\alpha} \\[5ex] = \dfrac{\sin\alpha - \cos\alpha}{\sin\alpha\cos\alpha} * \dfrac{\sin\alpha\cos\alpha}{1} \\[5ex] = \sin\alpha - \cos\alpha$
(57.) $\dfrac{\cos(\theta - \omega)}{\cos\theta\sin\omega}$

$\dfrac{\cos(\theta - \omega)}{\cos\theta\sin\omega} \\[5ex] \cos(\theta - \omega) = \cos\theta\cos\omega + \sin\theta\sin\omega ... Difference\:\: Formula \\[3ex] = \dfrac{\cos\theta\cos\omega + \sin\theta\sin\omega}{\cos\theta\sin\omega} \\[5ex] = \dfrac{\cos\theta\cos\omega}{\cos\theta\sin\omega} + \dfrac{\sin\theta\sin\omega}{\cos\theta\sin\omega} \\[5ex] = \dfrac{\cos\omega}{\sin\omega} + \dfrac{\sin\theta}{\cos\theta} \\[5ex] \dfrac{\cos\omega}{\sin\omega} = \cot\omega ... Quotient\:\: Identity \\[5ex] \dfrac{\sin\theta}{\cos\theta} = \tan\theta ... Quotient\:\: Identity \\[5ex] = \cot\omega + \tan\theta$
(58.) $\sec\alpha - \cos\alpha$

$\sec\alpha - \cos\alpha \\[3ex] \sec\alpha = \dfrac{1}{\cos\alpha} ... Reciprocal\:\: Identity \\[5ex] = \dfrac{1}{\cos\alpha} - \dfrac{\cos\alpha}{1} \\[5ex] = \dfrac{1 - \cos\alpha(\cos\alpha)}{\cos\alpha} \\[5ex] = \dfrac{1 - \cos^2\alpha}{\cos\alpha} \\[5ex] \sin^2\alpha + \cos^2\alpha = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \sin^2\alpha = 1 - \cos^2\alpha \\[3ex] = \dfrac{\sin^2\alpha}{\cos\alpha} \\[5ex] = \sin\alpha * \dfrac{\sin\alpha}{\cos\alpha} \\[5ex] \tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} ... Quotient\:\: Identity \\[5ex] = \sin\alpha * \tan\alpha \\[3ex] = \sin\alpha\tan\alpha$
(59.) $\dfrac{\cos(\theta - \omega)}{\cos\theta\cos\omega}$

$\dfrac{\cos(\theta - \omega)}{\cos\theta\cos\omega} \\[5ex] \cos(\theta - \omega) = \cos\theta\cos\omega + \sin\theta\sin\omega ... Difference\:\: Formula \\[3ex] = \dfrac{\cos\theta\cos\omega + \sin\theta\sin\omega}{\cos\theta\cos\omega} \\[5ex] = \dfrac{\cos\theta\cos\omega}{\cos\theta\cos\omega} + \dfrac{\sin\theta\sin\omega}{\cos\theta\cos\omega} \\[5ex] = 1 + \dfrac{\sin\theta\sin\omega}{\cos\theta\cos\theta} \\[5ex] \dfrac{\sin\theta}{\cos\theta} = \tan\theta ... Quotient\:\: Identity \\[5ex] \dfrac{\sin\omega}{\cos\omega} = \tan\omega ... Quotient\:\: Identity \\[5ex] = 1 + \tan\theta\tan\omega$
(60.) Rationalize the denominator

$\sqrt{\dfrac{\sin \theta}{\cos^7 \theta}}$

We want to rationalize the denominator
This means that we want to make the denominator a rational number
What do we have to multiply by $7$ (the exponent at the denominator is $7$) in terms of exponent so that it is divisible by $2$
Why $2$? ... it is because of the square root.
Remember: Square root means an exponent of $\dfrac{1}{2}$... division by $2$
$\sqrt{x} = x^\frac{1}{2}$
So, we have to multiply $\cos^7 \theta$ by $\cos \theta$ so we can get $\cos^8 \theta$ because $8$ is divisible by $2$
Whatever you do to the denominator, you have to do the same thing to the numerator
This is necessary in order to get an equivalent fraction to the original fraction
$\sqrt{\dfrac{\sin \theta}{\cos^7 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta}}{\sqrt{\cos^7 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta}}{\sqrt{\cos^7 \theta}} * \dfrac{\sqrt{\cos \theta}}{\sqrt{\cos \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta * \cos \theta}}{\sqrt{\cos^7 \theta * \cos \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta \cos \theta}}{\sqrt{\cos^8 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta \cos \theta}}{\cos^4 \theta} \\[5ex]$ The denominator is now a rational number.

(61.) $\dfrac{\tan^2\left(\dfrac{\omega}{2}\right) - 1}{\tan^2\left(\dfrac{\omega}{2}\right) + 1}$


(62.) ACT If $\cos x = -\dfrac{1}{3}$, what is the value of $\cos 2x$

$\left[Note:\:\: (\cos x)^2 = \dfrac{1 + \cos 2x}{2}\right]$

$(\cos x)^2 = \dfrac{1 + \cos 2x}{2} \\[5ex] Cross\:\:Multiply \\[3ex] 2 * \cos x * \cos x = 1 + \cos2x \\[3ex] 2\cos^2x = 1 + \cos2x \\[3ex] 1 + \cos2x = 2\cos^2x \\[3ex] \cos2x = 2 * \cos x * \cos x - 1 \\[3ex] \cos2x = 2 * -\dfrac{1}{3} * -\dfrac{1}{3} - 1 \\[5ex] \cos2x = \dfrac{2}{9} - \dfrac{9}{9} \\[5ex] \cos2x = \dfrac{2 - 9}{9} \\[5ex] \cos2x = -\dfrac{7}{9}$
(63.)


(64.) $\sin(-\theta) + \cos(-\theta)$

$\sin(-\theta) = -\sin \theta ... Odd\;\;Identity \\[3ex] \cos(-\theta) = \cos \theta ... Even\;\;Identity \\[3ex] \sin(-\theta) + \cos(-\theta) \\[3ex] = -\sin \theta + \cos \theta \\[3ex] = \cos \theta - \sin \theta$
(65.)


(66.)