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# Trigonometric Proofs / Prove Trigonometric Identities

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 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.

Prove each identity
State the reason(s) as applicable
Show all work

You may:
(1.) begin from the $LHS$ and end at the $RHS$ OR
(2.) begin from the $RHS$ and end at the $LHS$ OR
(3.) begin from the $LHS$ and end at a simplified answer, then begin from the $RHS$ and end at the same simplified answer. The third option should only be used when the previous two options are exhausted.
Use whichever side is easier to simplify.
Work towards getting the result of the other side.

(1.) $\tan \gamma * \cos \gamma = \sin \gamma$

$\underline{LHS} \\[3ex] \tan \gamma * \cos \gamma \\[3ex] \tan \gamma = \dfrac{\sin \gamma}{\cos \gamma} ...Quotient\:\: Identity \\[5ex] \rightarrow \dfrac{\sin \gamma}{\cos \gamma} * \cos \gamma \\[5ex] = \sin \gamma \\[3ex] = RHS$
(2.) $\sec \phi - \sin \phi \tan \phi = \cos \phi$

$\underline{LHS} \\[3ex] \sec \phi - \sin \phi \tan \phi \\[3ex] \sec \phi = \dfrac{1}{\cos \phi} ...Reciprocal\:\: Identity \\[5ex] \tan \phi = \dfrac{\sin \phi}{\cos \phi} ...Quotient\:\: Identity \\[5ex] \rightarrow \dfrac{1}{\cos \phi} - \sin \phi * \dfrac{\sin \phi}{\cos \phi} \\[5ex] = \dfrac{1}{\cos \phi} - \dfrac{\sin^2 \phi}{\cos \phi} \\[5ex] = \dfrac{1 - \sin^2 \phi}{\cos \phi} \\[5ex] \sin^2 \phi + \cos^2 \phi = 1 ...Pythagorean\:\: Identity \\[3ex] \cos^2 \phi = 1 - \sin^2 \phi \\[3ex] = \dfrac{\cos^2 \phi}{\cos \phi} \\[5ex] = \cos \phi \\[3ex] = RHS$
(3.) $\cos \alpha(\cot \alpha + \tan \alpha) = \csc \alpha$

$\underline{LHS} \\[3ex] \cos \alpha(\cot \alpha + \tan \alpha) \\[3ex] \cot \alpha = \dfrac{\cos \alpha}{\sin \alpha} ...Quotient\:\: Identity \\[5ex] \tan \alpha = \dfrac{\sin \alpha}{\cos \alpha} ...Quotient\:\: Identity \\[5ex] = \cos \alpha\left(\dfrac{\cos \alpha}{\sin \alpha} + \dfrac{\sin \alpha}{\cos \alpha}\right) \\[5ex] = \cos\alpha\left(\dfrac{\cos\alpha\cos\alpha + \sin\alpha\sin\alpha}{\sin\alpha\cos\alpha}\right) \\[5ex] = \dfrac{\cos^2\alpha + \sin^2\alpha}{\sin\alpha} \\[5ex] \sin^2 \alpha + \cos^2 \alpha = 1 ...Pythagorean\:\: Identity \\[3ex] = \dfrac{1}{\sin \alpha} \\[5ex] = \csc \alpha \\[3ex] = RHS$
(4.) $\tan \beta(\tan \beta + \cot \beta) = \sec^2 \beta$

$\underline{LHS} \\[3ex] \tan \beta(\tan \beta + \cot \beta) \\[3ex] \cot \beta = \dfrac{1}{\tan \beta} ... Reciprocal\:\: Identity \\[5ex] \rightarrow \tan \beta\left(\tan \beta + \dfrac{1}{\tan \beta}\right) \\[5ex] = \tan^2 \beta + 1 \\[3ex] = \sec^2 \beta ...Pythagorean\:\: Identity \\[3ex] = RHS$
(5.) $\sin4\omega + \sin6\omega = \cot\omega(\cos4\omega - \cos6\omega)$

$\underline{LHS} \\[3ex] \sin4\omega + \sin6\omega \\[3ex] = 2\sin\left(\dfrac{4\omega + 6\omega}{2}\right)\cos\left(\dfrac{4\omega - 6\omega}{2}\right) ...Sum-to-Product\:\: Formula \\[5ex] = 2\sin\left(\dfrac{10\omega}{2}\right)\cos\left(\dfrac{-2\omega}{2}\right) \\[5ex] = 2\sin(5\omega)\cos(-\omega) \\[3ex] \cos(-\omega) = \cos(\omega) ... Even\:\: Identity \\[3ex] = 2\sin5\omega\cos\omega ... Simplified\:\: Result \\[3ex] \underline{RHS} \\[3ex] \cot\omega(\cos4\omega - \cos6\omega) \\[3ex] \underline{Multiplicand} \\[3ex] \cos4\omega - \cos6\omega \\[3ex] = -2\sin\left(\dfrac{4\omega + 6\omega}{2}\right)\sin\left(\dfrac{4\omega - 6\omega}{2}\right) ...Sum-to-Product\:\: Formula \\[5ex] = -2\sin\left(\dfrac{10\omega}{2}\right)\sin\left(\dfrac{-2\omega}{2}\right) \\[5ex] = -2\sin(5\omega)\sin(-\omega) \\[3ex] \sin(-\omega) = -\sin(\omega) ... Odd\:\: Identity \\[3ex] = -2 * \sin5\omega * -\sin\omega \\[3ex] = 2\sin5\omega\sin\omega \\[3ex] \rightarrow RHS = \cot\omega * 2\sin5\omega\sin\omega \\[3ex] \cot\omega = \dfrac{\cos\omega}{\sin\omega} ... Quotient\:\: Identity \\[5ex] RHS = \dfrac{\cos\omega}{\sin\omega} * 2\sin5\omega\sin\omega \\[5ex] RHS = \cos\omega * 2\sin5\omega \\[3ex] RHS = 2\sin5\omega\cos\omega ... Same\:\: Simplified\:\: Result$
(6.) $\csc^2(-\alpha) - 1 = \cot^2 \alpha$

$\underline{LHS} \\[3ex] \csc^2(-\alpha) - 1 \\[3ex] = \csc(-\alpha) * \csc(-\alpha) - 1 \\[3ex] \csc(-\alpha) = -\csc \alpha ...Odd\:\: Identity \\[3ex] \rightarrow -\csc \alpha * -\csc \alpha - 1 \\[3ex] = \csc^2 \alpha - 1 \\[3ex] 1 + \cot^2 \alpha = \csc^2 \alpha ... Pythagorean\:\: Identity \\[3ex] \cot^2 \alpha = \csc^2 \alpha - 1 \\[3ex] = \cot^2 \alpha \\[3ex] = RHS$
(7.) $\dfrac{\csc \theta + 1}{\csc \theta - 1} + \dfrac{1 - \csc \theta}{\csc \theta + 1} = 4\sec \theta \tan \theta$

$\underline{LHS} \\[3ex] \dfrac{\csc \theta + 1}{\csc \theta - 1} + \dfrac{1 - \csc \theta}{\csc \theta + 1} \\[5ex] = \dfrac{(\csc \theta + 1)(\csc \theta + 1) + (1 - \csc \theta)(\csc \theta - 1)}{(\csc \theta - 1)(\csc \theta + 1)} \\[5ex] \underline{Numerator} \\[3ex] (\csc \theta + 1)(\csc \theta + 1) + (1 - \csc \theta)(\csc \theta - 1) \\[3ex] = \csc^2 \theta + 2\csc \theta + 1 + 2\csc \theta - 1 - \csc^2 \theta \\[3ex] = 4\csc \theta \\[3ex] \underline{Denominator} \\[3ex] (\csc \theta - 1)(\csc \theta + 1) = \csc^2 \theta - 1^2 ...Difference\:\: of\:\: Two\:\: Squares \\[3ex] = \csc^2 \theta - 1 \\[3ex] 1 + \cot^2 \theta = \csc^2 \theta ... Pythagorean\:\: Identity \\[3ex] \cot^2 \theta = \csc^2 \theta - 1 \\[3ex] = \cot^2 \theta \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{4\csc \theta}{\cot^2 \theta} \\[5ex] = 4\csc \theta \div \cot^2 \theta \\[3ex] \csc \theta = \dfrac{1}{\sin \theta} ... Reciprocal\:\: Identity \\[5ex] \cot \theta = \dfrac{\cos \theta}{\sin \theta} ... Quotient\:\: Identity \\[5ex] = 4 * \dfrac{1}{\sin \theta} \div \left(\dfrac{\cos \theta}{\sin \theta}\right)^2 \\[5ex] = \dfrac{4}{\sin \theta} \div \dfrac{\cos^2 \theta}{\sin^2 \theta} \\[5ex] = \dfrac{4}{\sin \theta} * \dfrac{\sin^2 \theta}{\cos^2 \theta} \\[5ex] = \dfrac{4\sin \theta}{\cos^2 \theta} \\[5ex] = 4 * \dfrac{\sin \theta}{\cos \theta} * \dfrac{1}{\cos \theta} \\[5ex] \dfrac{\sin \theta}{\cos \theta} = \tan \theta ... Quotient\:\: Identity \\[5ex] \dfrac{1}{\cos \theta} = \sec \theta ... Reciprocal\:\: Identity \\[5ex] = 4 * \tan \theta * \sec \theta \\[3ex] = 4\sec \theta \tan \theta \\[3ex] = RHS$
(8.) $2\sec \phi = \dfrac{1 - \sin \phi}{\cos \phi} + \dfrac{\cos \phi}{1 - \sin \phi}$

$\underline{RHS} \\[3ex] \dfrac{1 - \sin \phi}{\cos \phi} + \dfrac{\cos \phi}{1 - \sin \phi} \\[5ex] = \dfrac{(1 - \sin \phi)(1 - \sin \phi) + \cos \phi(\cos \phi)}{\cos \phi(1 - \sin \phi)} \\[5ex] \underline{Numerator} \\[3ex] (1 - \sin \phi)(1 - \sin \phi) + \cos \phi(\cos \phi) \\[3ex] = 1 - 2\sin \phi + \sin^2 \phi + \cos^2 \phi \\[3ex] \sin^2 \phi + \cos^2 \phi = 1 ...Pythagorean\:\: Identity \\[3ex] = 1 - 2\sin \phi + 1 \\[3ex] = 2 - 2\sin \phi \\[3ex] = 2(1 - \sin \phi) \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{2(1 - \sin \phi)}{\cos \phi(1 - \sin \phi)} \\[5ex] = \dfrac{2}{\cos \phi} \\[5ex] = 2 * \dfrac{1}{\cos \phi} \\[5ex] \dfrac{1}{\cos \phi} = \sec \phi ...Reciprocal\:\: Identity \\[5ex] = 2 * \sec \phi \\[3ex] = 2\sec \phi \\[3ex] = LHS$
(9.) $1 - \sin \beta \cos \beta = \dfrac{\sin^3 \beta + \cos^3 \beta}{\sin \beta + \cos \beta}$

$\underline{RHS} \\[3ex] \dfrac{\sin^3 \beta + \cos^3 \beta}{\sin \beta + \cos \beta} \\[5ex] \sin^3 \beta + \cos^3 \beta = (\sin \beta + \cos \beta)(\sin^2 \beta - \sin \beta \cos \beta + \cos^2 \beta) ...Sum \:\:of\:\: two\:\: cubes \\[3ex] = \dfrac{(\sin \beta + \cos \beta)(\sin^2 \beta - \sin \beta \cos \beta + \cos^2 \beta)}{\sin \beta + \cos \beta} \\[5ex] = \sin^2 \beta - \sin \beta \cos \beta + \cos^2 \beta \\[3ex] = \sin^2 \beta + \cos^2 \beta - \sin \beta \cos \beta \\[3ex] \sin^2 \beta + \cos^2 \beta = 1 ...Pythagorean\:\: Identity \\[3ex] = 1 - \sin \beta \cos \beta \\[3ex] = LHS$
(10.) $\dfrac{\csc \alpha - 1}{\csc \alpha + 1} = \dfrac{1 - \sin \alpha}{1 + \sin \alpha}$

$\underline{LHS} \\[3ex] \dfrac{\csc \alpha - 1}{\csc \alpha + 1} \\[5ex] \underline{Numerator} \\[3ex] \csc \alpha - 1 \\[3ex] \csc \alpha = \dfrac{1}{\sin \alpha} ... Reciprocal\:\: Identity \\[5ex] = \dfrac{1}{\sin \alpha} - 1 \\[5ex] = \dfrac{1}{\sin \alpha} - \dfrac{\sin \alpha}{\sin \alpha} \\[5ex] = \dfrac{1 - \sin \alpha}{\sin \alpha} \\[5ex] \underline{Denominator} \\[3ex] \csc \alpha + 1 \\[3ex] = \dfrac{1}{\sin \alpha} + 1 \\[5ex] = \dfrac{1}{\sin \alpha} + \dfrac{\sin \alpha}{\sin \alpha} \\[5ex] = \dfrac{1 + \sin \alpha}{\sin \alpha} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{1 - \sin \alpha}{\sin \alpha} \div \dfrac{1 + \sin \alpha}{\sin \alpha} \\[5ex] = \dfrac{1 - \sin \alpha}{\sin \alpha} * \dfrac{\sin \alpha}{1 + \sin \alpha} \\[5ex] = \dfrac{1 - \sin \alpha}{1 + \sin \alpha} \\[5ex] = RHS$
(11.) $\sin(x + y) \sin(x - y) = \sin^2 x - \sin^2 y$

$\underline{LHS} \\[3ex] \sin(x + y) \sin(x - y) \\[3ex] \sin(x + y) = \sin x \cos y + \cos x \sin y ...Sum\:\: Identity \\[3ex] \sin(x - y) = \sin x \cos y - \cos x \sin y ...Difference\:\: Identity \\[3ex] = (\sin x \cos y + \cos x \sin y)(\sin x \cos y - \cos x \sin y) \\[3ex] = \sin^2 x \cos^2 y - \sin x \cos x \sin y \cos y + \sin x \cos x \sin y \cos y - \cos^2 x \sin^2 y \\[3ex] = \sin^2 x \cos^2 y - \cos^2 x \sin^2 y \\[3ex] \sin^2 x + \cos^2 x = 1 ... Pythagorean\:\: Identity \\[3ex] \cos^2 x = 1 - \sin^2 x \\[3ex] \sin^2 y + \cos^2 y = 1 ... Pythagorean\:\: Identity \\[3ex] \cos^2 y = 1 - \sin^2 y \\[3ex] = \sin^2 x(1 - \sin^2 y) - (1 - \sin^2 x)(\sin^2 y) \\[3ex] = \sin^2 x(1 - \sin^2 y) - \sin^2 y(1 - \sin^2 x) \\[3ex] = \sin^2 x - \sin^2 x \sin^2 y - \sin^2 y + \sin^2 y \sin^2 x \\[3ex] = \sin^2 x - \sin^2 y - \sin^2 x \sin^2 y + \sin^2 x \sin^2 y \\[3ex] = \sin^2 x - \sin^2 y \\[3ex] = RHS$
(12.) $\sin \beta = 1 - \dfrac{\cos^2 \beta}{1 + \sin \beta}$

$\underline{RHS} \\[3ex] 1 - \dfrac{\cos^2 \beta}{1 + \sin \beta} \\[5ex] = \dfrac{1 + \sin \beta}{1 + \sin \beta} - \dfrac{\cos^2 \beta}{1 + \sin \beta} \\[5ex] \dfrac{1 + \sin \beta - \cos^2 \beta}{1 + \sin \beta} \\[5ex] \underline{Numerator} \\[3ex] 1 + \sin \beta - \cos^2 \beta \\[3ex] \sin^2 \beta + \cos^2 \beta = 1 ... Pythagorean\:\: Identity \\[3ex] \cos^2 \beta = 1 - \sin^2 \beta \\[3ex] = 1 + \sin \beta - (1 - \sin^2 \beta) \\[3ex] = 1 + \sin \beta - 1 + \sin^2 \beta \\[3ex] = \sin \beta + \sin^2 \beta \\[3ex] = \sin \beta(1 + \sin \beta) \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{\sin \beta(1 + \sin \beta)}{1 + \sin \beta} \\[5ex] = \sin \beta \\[3ex] = LHS$
(13.) $\cos^2 \phi = \dfrac{\cos^2 \phi - \sin^2 \phi}{1 - \tan^2 \phi}$

$\underline{RHS} \\[3ex] \dfrac{\cos^2 \phi - \sin^2 \phi}{1 - \tan^2 \phi} \\[5ex] \underline{Denominator} \\[3ex] 1 - \tan^2 \phi \\[3ex] \tan \phi = \dfrac{\sin \phi}{\cos \phi} ... Quotient\:\: Identity \\[5ex] \tan^2 \phi = \dfrac{\sin^2 \phi}{\cos^2 \phi} \\[5ex] = 1 - \dfrac{\sin^2 \phi}{\cos^2 \phi} \\[5ex] = \dfrac{\cos^2 \phi}{\cos^2 \phi} - \dfrac{\sin^2 \phi}{\cos^2 \phi} \\[5ex] = \dfrac{\cos^2 \phi - \sin^2 \phi}{\cos^2 \phi} \\[5ex] \underline{RHS} \\[3ex] \dfrac{\cos^2 \phi - \sin^2 \phi}{1 - \tan^2 \phi} \\[5ex] = (\cos^2 \phi - \sin^2 \phi) \div \dfrac{\cos^2 \phi - \sin^2 \phi}{\cos^2 \phi} \\[5ex] = (\cos^2 \phi - \sin^2 \phi) * \dfrac{\cos^2 \phi}{\cos^2 \phi - \sin^2 \phi} \\[5ex] = \cos^2 \phi \\[3ex] = LHS$
(14.) $\dfrac{\sec \theta - 1}{\tan \theta} = \dfrac{\tan \theta}{\sec \theta + 1}$

This is a bit tricky.
Let us try to simplify both sides and see if we can get the same simplified result.

$\underline{LHS} \\[3ex] \dfrac{\sec \theta - 1}{\tan \theta} \\[5ex] \underline{Numerator} \\[3ex] \sec \theta - 1 \\[3ex] \sec \theta = \dfrac{1}{\cos \theta} ... Reciprocal\:\: Identity \\[5ex] = \dfrac{1}{\cos \theta} - 1 \\[5ex] = \dfrac{1}{\cos \theta} - \dfrac{\cos \theta}{\cos \theta} \\[5ex] = \dfrac{1 - \cos \theta}{\cos \theta} \\[5ex] \underline{Denominator} \\[3ex] \tan \theta \\[3ex] \tan \theta = \dfrac{\sin \theta}{\cos \theta} ... Quotient\:\: Identity \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{1 - \cos \theta}{\cos \theta} \div \dfrac{\sin \theta}{\cos \theta} \\[5ex] = \dfrac{1 - \cos \theta}{\cos \theta} * \dfrac{\cos \theta}{\sin \theta} \\[5ex] = \dfrac{1 - \cos \theta}{\sin \theta} ... simplified\:\: result \\[5ex] \underline{RHS} \\[3ex] \dfrac{\tan \theta}{\sec \theta + 1} \\[5ex] \underline{Denominator} \\[3ex] \sec \theta + 1 = \dfrac{1}{\cos \theta} + 1 \\[5ex] = \dfrac{1}{\cos \theta} + \dfrac{\cos \theta}{\cos \theta} \\[5ex] = \dfrac{1 + \cos \theta}{\cos \theta} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \tan \theta \div (\sec \theta + 1) \\[3ex] = \dfrac{\sin \theta}{\cos \theta} \div \dfrac{1 + \cos \theta}{\cos \theta} \\[5ex] = \dfrac{\sin \theta}{\cos \theta} * \dfrac{\cos \theta}{1 + \cos \theta} \\[5ex] = \dfrac{\sin \theta}{1 + \cos \theta} ... simplified\:\: result \\[5ex]$ But, the simplified results are not the same.
So, we have to come up with another technique.
Let us think about multiplying the numerator and denominator by "something".
We just need to get the other side.
Let us multiply the numerator and denominator of the $LHS$ by $\sec \theta + 1$

$\underline{LHS} \\[3ex] \dfrac{\sec \theta - 1}{\tan \theta} \\[5ex] \underline{New\:\: LHS} \\[3ex] \dfrac{(\sec \theta + 1)(\sec \theta - 1)}{(\sec \theta + 1)(\tan \theta)} \\[5ex] \underline{Numerator} \\[3ex] (\sec \theta + 1)(\sec \theta - 1) = \sec^2 \theta - 1^2 ... Diff. \:\:of\:\: Two\:\: Squares \\[3ex] = \sec^2 \theta - 1 \\[3ex] 1 + \tan^2 \theta = \sec^2 \theta ... Pythagorean\:\: Identity \\[3ex] \tan^2 \theta = \sec^2 \theta - 1 \\[3ex] = \tan^2 \theta \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{\tan^2 \theta}{(\sec \theta + 1)(\tan \theta)} \\[5ex] = \dfrac{\tan \theta}{\sec \theta + 1} \\[5ex] = RHS$
(15.) $\dfrac{1 - \sin\theta}{1 + \sin\theta} = (\sec\theta - \tan\theta)^2$

$\underline{RHS} \\[3ex] (\sec\theta - \tan\theta)^2 \\[3ex] = (\sec\theta - \tan\theta)(\sec\theta - \tan\theta) \\[3ex] = \sec^2\theta - \tan\theta\sec\theta - \tan\theta\sec\theta + \tan^2\theta \\[3ex] = \sec^2\theta - 2\tan\theta\sec\theta + \tan^2\theta \\[3ex] \tan\theta = \dfrac{\sin\theta}{\cos\theta} ...Quotient\:\: Identity \\[5ex] \sec\theta = \dfrac{1}{\cos\theta} ...Reciprocal\:\: Identity \\[5ex] \sec^c\theta = \left(\dfrac{1}{\cos\theta}\right)^2 = \dfrac{1^2}{\cos^2\theta} = \dfrac{1}{\cos^2\theta} \\[5ex] 2\tan\theta\sec\theta = 2 * \dfrac{\sin\theta}{\cos\theta} * \dfrac{1}{\cos\theta} = \dfrac{2\sin\theta}{\cos^2\theta} \\[5ex] \tan^2\theta = \left(\dfrac{\sin\theta}{\cos\theta}\right)^2 = \dfrac{\sin^2\theta}{\cos^2\theta} \\[5ex] = \dfrac{1}{\cos^2\theta} - \dfrac{2\sin\theta}{\cos^2\theta} + \dfrac{\sin^2\theta}{\cos^2\theta} \\[5ex] = \dfrac{1 - 2\sin\theta + \sin^2\theta}{\cos^2\theta} \\[5ex] \underline{Numerator} \\[3ex] 1 - 2\sin\theta + \sin^2\theta \\[3ex] = \sin^2\theta - 2\sin\theta + 1 ...Quadratic\:\: in\:\: \sin\theta \\[3ex] Let\:\: p = \sin\theta \\[3ex] = p^2 - 2p + 1 \\[3ex] = (p - 1)(p - 1) \\[3ex] Substitute\:\: back \\[3ex] = (\sin\theta - 1)(\sin\theta - 1) \\[3ex] \underline{Denominator} \\[3ex] \cos^2\theta \\[3ex] \sin^2\theta + \cos^2\theta = 1 ... Pythagorean\:\: Idenitity \\[3ex] = 1 - \sin^2\theta \\[3ex] = 1^2 - \sin^2\theta = (1 + \sin\theta)(1 - \sin\theta) ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] \dfrac{(\sin\theta - 1)(\sin\theta - 1)}{(1 + \sin\theta)(1 - \sin\theta)} \\[5ex]$ We need to get to the $LHS$
We have to do some acceptable arithmetic operations/manipulations ☺
Multiply both the numerator and the denominator by $-1$

$\underline{Numerator} \\[3ex] (\sin\theta - 1)(\sin\theta - 1) \\[3ex] -1 * (\sin\theta - 1) * (\sin\theta - 1) \\[3ex] = (-\sin\theta + 1)(\sin\theta - 1) \\[3ex] = (1 - \sin\theta)(\sin\theta - 1) \\[3ex] \underline{Denominator} \\[3ex] (1 + \sin\theta)(1 - \sin\theta) \\[3ex] -1 * (1 + \sin\theta) * (1 - \sin\theta) \\[3ex] = -1 * (1 - \sin\theta) * (1 + \sin\theta) \\[3ex] = (-1 + \sin\theta)(1 + \sin\theta) \\[3ex] = (\sin\theta - 1)(1 + \sin\theta) \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] \dfrac{(1 - \sin\theta)(\sin\theta - 1)}{(\sin\theta - 1)(1 + \sin\theta)} \\[5ex] = \dfrac{1 - \sin\theta}{1 + \sin\theta} \\[5ex] = LHS$
(16.) $\dfrac{2\tan\theta}{1 + \tan^2\theta} = \sin2\theta$

$\underline{LHS} \\[3ex] \dfrac{2\tan\theta}{1 + \tan^2\theta} \\[5ex] \underline{Numerator} \\[3ex] 2\tan\theta \\[3ex] \tan\theta = \dfrac{\sin\theta}{\cos\theta} ... Quotient\:\: Identity \\[5ex] = 2 * \dfrac{\sin\theta}{\cos\theta} = \dfrac{2\sin\theta}{\cos\theta} \\[5ex] \underline{Denominator} \\[3ex] 1 + \tan^2\theta \\[3ex] = \sec^2\theta ... Pythagorean\:\: Identity \\[3ex] \sec\theta = \dfrac{1}{\cos\theta} ... Reciprocal\:\: Identity \\[5ex] \sec^2\theta = \left(\dfrac{1}{\cos\theta}\right)^2 = \dfrac{1^2}{\cos^2\theta} = \dfrac{1}{\cos^2\theta} \\[5ex] = \dfrac{1}{\cos^2\theta} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{2\sin\theta}{\cos\theta} \div \dfrac{1}{\cos^2\theta} \\[5ex] = \dfrac{2\sin\theta}{\cos\theta} * \dfrac{\cos^2\theta}{1} \\[5ex] = 2\sin\theta\cos\theta \\[3ex] = \sin2\theta ... Double-Angle\:\: Formula \\[3ex] = RHS$
(17.) $\dfrac{1 + \sin\theta}{1 - \sin\theta} = (\sec\theta + \tan\theta)^2$

$\underline{RHS} \\[3ex] (\sec\theta + \tan\theta)^2 \\[3ex] = (\sec\theta + \tan\theta)(\sec\theta + \tan\theta) \\[3ex] = \sec^2\theta + \tan\theta\sec\theta + \tan\theta\sec\theta + \tan^2\theta \\[3ex] = \sec^2\theta + 2\tan\theta\sec\theta + \tan^2\theta \\[3ex] \tan\theta = \dfrac{\sin\theta}{\cos\theta} ...Quotient\:\: Identity \\[5ex] \sec\theta = \dfrac{1}{\cos\theta} ...Reciprocal\:\: Identity \\[5ex] \sec^c\theta = \left(\dfrac{1}{\cos\theta}\right)^2 = \dfrac{1^2}{\cos^2\theta} = \dfrac{1}{\cos^2\theta} \\[5ex] 2\tan\theta\sec\theta = 2 * \dfrac{\sin\theta}{\cos\theta} * \dfrac{1}{\cos\theta} = \dfrac{2\sin\theta}{\cos^2\theta} \\[5ex] \tan^2\theta = \left(\dfrac{\sin\theta}{\cos\theta}\right)^2 = \dfrac{\sin^2\theta}{\cos^2\theta} \\[5ex] = \dfrac{1}{\cos^2\theta} + \dfrac{2\sin\theta}{\cos^2\theta} + \dfrac{\sin^2\theta}{\cos^2\theta} \\[5ex] = \dfrac{1 + 2\sin\theta + \sin^2\theta}{\cos^2\theta} \\[5ex] \underline{Numerator} \\[3ex] 1 + 2\sin\theta + \sin^2\theta \\[3ex] = \sin^2\theta - 2\sin\theta + 1 ...Quadratic\:\: in\:\: \sin\theta \\[3ex] Let\:\: p = \sin\theta \\[3ex] = p^2 + 2p + 1 \\[3ex] = (p + 1)(p + 1) \\[3ex] Substitute\:\: back \\[3ex] = (\sin\theta + 1)(\sin\theta + 1) \\[3ex] = (1 + \sin\theta)(1 + \sin\theta) \\[3ex] \underline{Denominator} \\[3ex] \cos^2\theta \\[3ex] \sin^2\theta + \cos^2\theta = 1 ... Pythagorean\:\: Idenitity \\[3ex] = 1 - \sin^2\theta \\[3ex] = 1^2 - \sin^2\theta = (1 + \sin\theta)(1 - \sin\theta) ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] \dfrac{(1 + \sin\theta)(1 + \sin\theta)}{(1 + \sin\theta)(1 - \sin\theta)} \\[5ex] = \dfrac{1 + \sin\theta}{1 - \sin\theta} \\[5ex] = LHS$
(18.) $\tan\theta + \cot\theta = \dfrac{\tan\theta}{\sin^2\theta}$

$\underline{RHS} \\[3ex] \dfrac{\tan\theta}{\sin^2\theta} \\[5ex] \tan\theta = \dfrac{\sin\theta}{\cos\theta} ... Quotient\:\: Identity \\[5ex] = \dfrac{\sin\theta}{\cos\theta} \div \sin^2\theta \\[5ex] = \dfrac{\sin\theta}{\cos\theta} * \dfrac{1}{\sin^2\theta} \\[5ex] = \dfrac{1}{\cos\theta\sin\theta} \\[5ex] = \dfrac{1}{\cos\theta} * \dfrac{1}{\sin\theta} \\[5ex] = \sec\theta\csc\theta...simplified\:\: result \\[3ex] \underline{LHS} \\[3ex] \tan\theta + \cot\theta \\[3ex] \cot\theta = \dfrac{\cos\theta}{\sin\theta} ... Quotient\:\: Identity \\[5ex] = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} \\[5ex] = \dfrac{\sin\theta(\sin\theta) + \cos\theta(\cos\theta)}{\cos\theta\sin\theta} \\[5ex] = \dfrac{\sin^2\theta + \cos^2\theta}{\cos\theta\sin\theta} \\[5ex] \sin^2\theta + \cos^2\theta = 1 ... Pythagorean\:\: Identity \\[3ex] = \dfrac{1}{\cos\theta\sin\theta} \\[5ex] = \dfrac{1}{\cos\theta} * \dfrac{1}{\sin\theta} \\[5ex] = \sec\theta\csc\theta...same\:\: simplified\:\: result$
(19.) $\cos^4\theta - \sin^4\theta = \cos2\theta$

$\underline{LHS} \\[3ex] \cos^4\theta - \sin^4\theta \\[3ex] = (\cos^2\theta)^2 - (\sin^2\theta)^2 \\[3ex] = (\cos^2\theta + \sin^2\theta)(\cos^2\theta - \sin^2\theta) ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] \cos^2\theta + \sin^2\theta = 1 ... Pythagorean\:\: Identity \\[3ex] = 1 * \cos^2\theta - \sin^2\theta \\[3ex] = \cos^2\theta - \sin^2\theta \\[3ex] = \cos2\theta ... Double-Angle\:\: Identity \\[3ex] = RHS$
(20.) $\sin^2\theta\tan^2\theta = \tan^2\theta - \sin^2\theta$

$\underline{LHS} \\[3ex] \sin^2\theta\tan^2\theta \\[3ex] \sin^2\theta + \cos^2\theta = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \sin^2\theta = 1 - \cos^2\theta \\[3ex] = (1 - \cos^2\theta) * \tan^2\theta \\[3ex] = \tan^2\theta(1 - \cos^2\theta) \\[3ex] = \tan^2\theta - \tan^2\theta\cos^2\theta \\[3ex] \tan\theta = \dfrac{\sin\theta}{\cos\theta} ... Quotient\:\: Identity \\[5ex] \tan^2\theta = \left(\dfrac{\sin\theta}{\cos\theta}\right)^2 = \dfrac{\sin^2\theta}{\cos^2\theta} \\[5ex] = \tan^2\theta - \dfrac{\sin^2\theta}{\cos^2\theta} * \cos^2\theta \\[5ex] = \tan^2\theta - \sin^2\theta \\[3ex] = RHS$

(21.) If $\tan\alpha = p + 1$ and $\tan\beta = p - 1$, show that $2\cot(\alpha - \beta) = p^2$

$\tan\alpha = p + 1 \\[3ex] \tan\beta = p - 1 \\[3ex] Show\:\: that\:\: 2\cot(\alpha - \beta) = p^2 \\[3ex] \underline{LHS} \\[3ex] 2\cot(\alpha - \beta) \\[3ex] \cot(\alpha - \beta) = \dfrac{1}{\tan(\alpha - \beta)} ... Reciprocal\:\: Identity \\[5ex] \tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} ... Difference\:\: Formula \\[5ex] \tan(\alpha - \beta) = \dfrac{(p + 1) - (p - 1)}{1 + [(p + 1)(p - 1)]} \\[5ex] (p + 1)(p - 1) = p^2 - 1^2 ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] \tan(\alpha - \beta) = \dfrac{p + 1 - p + 1}{1 + (p^2 - 1^2)} \\[5ex] \tan(\alpha - \beta) = \dfrac{2}{1 + p^2 - 1} \\[5ex] \tan(\alpha - \beta) = \dfrac{2}{p^2} \\[5ex] \dfrac{1}{\tan(\alpha - \beta)} = 1 \div \tan(\alpha - \beta) \\[3ex] \dfrac{1}{\tan(\alpha - \beta)} = 1 \div \dfrac{2}{p^2} \\[5ex] \dfrac{1}{\tan(\alpha - \beta)} = 1 * \dfrac{p^2}{2} \\[5ex] \dfrac{1}{\tan(\alpha - \beta)} = \dfrac{p^2}{2} \\[5ex] = 2 * \cot(\alpha - \beta) \\[3ex] = 2 * \dfrac{1}{\tan(\alpha - \beta)} \\[5ex] = 2 * \dfrac{p^2}{2} \\[5ex] = p^2 \\[3ex] = RHS$
(22.) $\dfrac{\tan\beta}{\tan\beta - 1} = \dfrac{1}{1 - \cot\beta}$

$\underline{RHS} \\[3ex] \dfrac{1}{1 - \cot\beta} \\[5ex] \underline{Denominator} \\[3ex] 1 - \cot\beta \\[3ex] \cot\beta = \dfrac{1}{\tan\beta} ... Reciprocal\:\: Identity \\[5ex] = 1 - \dfrac{1}{\tan\beta} \\[5ex] = \dfrac{\tan\beta}{\tan\beta} - \dfrac{1}{\tan\beta} \\[5ex] = \dfrac{\tan\beta - 1}{\tan\beta} \\[5ex] \rightarrow \dfrac{1}{1 - \cot\beta} \\[5ex] = 1 \div (1 - \cot\beta) \\[3ex] = 1 \div \dfrac{\tan\beta - 1}{\tan\beta} \\[5ex] = 1 * \dfrac{\tan\beta}{\tan\beta - 1} \\[5ex] = \dfrac{\tan\beta}{\tan\beta - 1} \\[5ex] = LHS$
(23.) $1 - \sin^4\theta = 2\sin^2\theta\cos^2\theta + \cos^4\theta$

$\underline{RHS} \\[3ex] 2\sin^2\theta\cos^2\theta + \cos^4\theta \\[3ex] = \cos^2\theta(2\sin^2\theta + \cos^2\theta) \\[3ex] \sin^2\theta + \cos^2\theta = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \cos^2\theta = 1 - \sin^2\theta \\[3ex] = (1 - \sin^2\theta)[2\sin^2\theta + (1 - \sin^2\theta)] \\[3ex] = (1 - \sin^2\theta)(2\sin^2\theta + 1 - \sin^2\theta) \\[3ex] = (1 - \sin^2\theta)(1 + \sin^2\theta) \\[3ex] = 1^2 - (\sin^2\theta)^2 ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] = 1 - \sin^4\theta \\[3ex] = LHS$
(24.) $-\cot\theta + \dfrac{\sin\theta}{1 - \cos\theta} = \csc\theta$

$\underline{LHS} \\[3ex] -\cot\theta + \dfrac{\sin\theta}{1 - \cos\theta} \\[5ex] \cot\theta = \dfrac{\cos\theta}{\sin\theta} \\[5ex] = \dfrac{\cos\theta}{\sin\theta} + \dfrac{\sin\theta}{1 - \cos\theta} \\[5ex] = \dfrac{-\cos\theta(1 - \cos\theta) + \sin\theta(\sin\theta)}{\sin\theta(1 - \cos\theta)} \\[5ex] = \dfrac{-\cos\theta + \cos^2\theta + \sin^2\theta}{\sin\theta(1 - \cos\theta)} \\[5ex] \cos^2\theta + \sin^2\theta = 1 ... Pythagorean\:\: Identity \\[3ex] = \dfrac{-\cos\theta + 1}{\sin\theta(1 - \cos\theta)} \\[5ex] = \dfrac{1 - \cos\theta}{\sin\theta(1 - \cos\theta)} \\[5ex] = \dfrac{1}{\sin\theta} \\[5ex] = \csc\theta ... Reciprocal\:\: Identity \\[3ex] = RHS$
(25.) $\log(\cos2\theta) = \log(\cos\theta - \sin\theta) + \log(\cos\theta + \sin\theta)$

$\underline{RHS} \\[3ex] \log(\cos\theta - \sin\theta) + \log(\cos\theta + \sin\theta) \\[3ex] = \log[(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)] ...Law\:\: 1...Log \\[3ex] (\cos\theta - \sin\theta)(\cos\theta + \sin\theta) = \cos^2\theta - \sin^2\theta ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] \cos^2\theta - \sin^2\theta = \cos2\theta ... Double-Angle\:\: Formula \\[3ex] = \log(\cos2\theta) \\[3ex] = LHS$
(26.) $\ln|\sec\theta| = -\ln|\cos\theta|$

$\underline{LHS} \\[3ex] \ln|\sec\theta| \\[3ex] = \dfrac{1}{\ln|\cos\theta|} ... Reciprocal\:\: Identity \\[5ex] = \left(\ln|\cos\theta|\right)^{-1} ... Law\:\: 6...Exp \\[3ex] = -1 * \ln|\cos\theta| ... Law\:\: 5 ...Log \\[3ex] = -\ln|\cos\theta| \\[3ex] = RHS$
(27.) $\ln|1 + \cos\beta| + \ln|1 - \cos\beta| = 2\ln|\sin\beta|$

$\underline{LHS} \\[3ex] \ln|1 + \cos\beta| + \ln|1 - \cos\beta| \\[3ex] = \ln\left[|1 + \cos\beta| * |1 - \cos\beta|\right] ...Law\:\: 1...Log \\[3ex] |1 + \cos\beta| * |1 - \cos\beta| = |(1 + \cos\beta) * (1 - \cos\beta)| \\[3ex] (1 + \cos\beta) * (1 - \cos\beta) = 1^2 - \cos^2\beta = 1 - \cos^2\beta ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] \sin^2\beta + \cos^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] \sin^2\beta = 1 - \cos^2\beta \\[3ex] \rightarrow |(1 + \cos\beta) * (1 - \cos\beta)| = |\sin^2\beta| \\[3ex] |\sin^2\beta| = |\sin\beta|^2 \\[3ex] = \ln|\sin\beta|^2 \\[3ex] = 2\ln|\sin\beta| ... Law\:\:5...Log \\[3ex] = RHS$
(28.) $\dfrac{\cot\beta - 1}{\cot\beta + 1} = \dfrac{1 - \tan\beta}{1 + \tan\beta}$

$\underline{LHS} \\[3ex] \dfrac{\cot\beta - 1}{\cot\beta + 1} \\[5ex] \cot\beta = \dfrac{\cos\beta}{\sin\beta} ... Quotient\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] \cot\beta - 1 \\[3ex] = \dfrac{\cos\beta}{\sin\beta} - 1 \\[5ex] = \dfrac{\cos\beta}{\sin\beta} - \dfrac{\sin\beta}{\sin\beta} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\sin\beta} \\[5ex] \underline{Denominator} \\[3ex] \cot\beta + 1 \\[3ex] = \dfrac{\cos\beta}{\sin\beta} + 1 \\[5ex] = \dfrac{\cos\beta}{\sin\beta} + \dfrac{\sin\beta}{\sin\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\sin\beta} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\sin\beta} \div \dfrac{\cos\beta + \sin\beta}{\sin\beta} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\sin\beta} * \dfrac{\sin\beta}{\cos\beta + \sin\beta} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\cos\beta + \sin\beta} ...Simplified\:\: Result \\[5ex] \underline{RHS} \\[3ex] \dfrac{1 - \tan\beta}{1 + \tan\beta} \\[5ex] \tan\beta = \dfrac{\sin\beta}{\cos\beta} ... Quotient\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] 1 - \tan\beta \\[3ex] = 1 - \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta}{\cos\beta} - \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\cos\beta} \\[5ex] \underline{Denominator} \\[3ex] 1 + \tan\beta \\[3ex] = 1 + \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta}{\cos\beta} + \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\cos\beta} \div \dfrac{\cos\beta + \sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\cos\beta} * \dfrac{\cos\beta}{\cos\beta + \sin\beta} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\cos\beta + \sin\beta} ...Same\:\: Simplified\:\: Result$
(29.) $5\cos^2\omega = \dfrac{4\cot\omega - \tan\omega}{\cot\omega + \tan\omega} + 1$

$\underline{RHS} \\[3ex] \dfrac{4\cot\omega - \tan\omega}{\cot\omega + \tan\omega} + 1 \\[5ex] \underline{Augend} \\[3ex] \dfrac{4\cot\omega - \tan\omega}{\cot\omega + \tan\omega} \\[5ex] \cot\omega = \dfrac{1}{\tan\omega} ... Reciprocal\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] 4\cot\omega - \tan\omega \\[3ex] = 4\left(\dfrac{1}{\tan\omega}\right) - \tan\omega \\[5ex] = \dfrac{4}{\tan\omega} - \dfrac{\tan\omega}{1} \\[5ex] = \dfrac{4 - \tan\omega(\tan\omega)}{\tan\omega} \\[5ex] = \dfrac{4 - \tan^2\omega}{\tan\omega} \\[5ex] \underline{Denominator} \\[3ex] \cot\omega + \tan\omega \\[3ex] = \dfrac{1}{\tan\omega} + \tan\omega \\[5ex] = \dfrac{1}{\tan\omega} + \dfrac{\tan\omega}{1} \\[5ex] = \dfrac{1 + \tan\omega(\tan\omega)}{\tan\omega} \\[5ex] = \dfrac{1 + \tan^2\omega}{\tan\omega} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{4 - \tan^2\omega}{\tan\omega} \div \dfrac{1 + \tan^2\omega}{\tan\omega} \\[5ex] = \dfrac{4 - \tan^2\omega}{\tan\omega} * \dfrac{\tan\omega}{1 + \tan^2\omega} \\[5ex] = \dfrac{4 - \tan^2\omega}{1 + \tan^2\omega} \\[5ex] 1 + \tan^2\omega = \sec^2\omega ... Pythagorean\:\: Identity \\[3ex] = \dfrac{4 - \tan^2\omega}{\sec^2\omega} \\[5ex] = \dfrac{4}{\sec^2\omega} - \dfrac{\tan^2\omega}{\sec^2\omega} \\[5ex] \sec\omega = \dfrac{1}{\cos\omega} ... Reciprocal\:\: Identity \\[5ex] \therefore \sec^2\omega = \dfrac{1^2}{\cos^2\omega} = \dfrac{1}{\cos^2\omega} \\[5ex] \tan\omega = \dfrac{\sin\omega}{\cos\omega} ... Quotient\:\: Identity \\[5ex] \therefore \tan^2\omega = \dfrac{\sin^2\omega}{\cos^2\omega} \\[5ex] \dfrac{4}{\sec^2\omega} = 4 \div \sec^2\omega = 4 \div \dfrac{1}{\cos^2\omega} = 4 * \dfrac{\cos^2\omega}{1} = 4\cos^2\omega \\[5ex] \dfrac{\tan^2\omega}{\sec^2\omega} = \tan^2\omega \div \sec^2\omega = \dfrac{\sin^2\omega}{\cos^2\omega} \div \dfrac{1}{\cos^2\omega} = \dfrac{\sin^2\omega}{\cos^2\omega} * \dfrac{\cos^2\omega}{1} = \sin^2\omega \\[5ex] = 4\cos^2\omega - \sin^2\omega \\[3ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \sin^2\omega = 1 - \cos^2\omega \\[3ex] = 4\cos^2\omega - (1 - \cos^2\omega) \\[3ex] = 4\cos^2\omega - 1 + \cos^2\omega \\[3ex] = 5\cos^2\omega - 1 \\[3ex] \rightarrow RHS = 5\cos^2\omega - 1 + 1 \\[3ex] = 5\cos^2\omega \\[3ex] = LHS$
(30.) $\dfrac{\cos^2\theta + \cot\theta}{\cos^2\theta - \cot\theta} = \dfrac{\cos^2\theta\tan\theta + 1}{\cos^2\theta\tan\theta - 1}$

$\underline{LHS} \\[3ex] \dfrac{\cos^2\theta + \cot\theta}{\cos^2\theta - \cot\theta} \\[5ex] \cot\theta = \dfrac{\cos\theta}{\sin\theta} ... Quotient\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] \cos^2\theta + \cot\theta \\[3ex] = \dfrac{\cos^2\theta}{1} + \dfrac{\cos\theta}{\sin\theta} \\[5ex] = \dfrac{\cos^2\theta\sin\theta + \cos\theta}{\sin\theta} \\[5ex] = \dfrac{\cos\theta(\cos\theta\sin\theta + 1)}{\sin\theta} \\[5ex] \underline{Denominator} \\[3ex] \cos^2\theta - \cot\theta \\[3ex] = \dfrac{\cos^2\theta}{1} - \dfrac{\cos\theta}{\sin\theta} \\[5ex] = \dfrac{\cos^2\theta\sin\theta - \cos\theta}{\sin\theta} \\[5ex] = \dfrac{\cos\theta(\cos\theta\sin\theta - 1)}{\sin\theta} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{\cos\theta(\cos\theta\sin\theta + 1)}{\sin\theta} \div \dfrac{\cos\theta(\cos\theta\sin\theta - 1)}{\sin\theta} \\[5ex] = \dfrac{\cos\theta(\cos\theta\sin\theta + 1)}{\sin\theta} * \dfrac{\sin\theta}{\cos\theta(\cos\theta\sin\theta - 1)} \\[5ex] = \dfrac{\cos\theta\sin\theta + 1}{\cos\theta\sin\theta - 1} ... Simplified\:\: Result \\[5ex] \underline{RHS} \\[3ex] \dfrac{\cos^2\theta\tan\theta + 1}{\cos^2\theta\tan\theta - 1} \\[5ex] \tan\theta = \dfrac{\sin\theta}{\cos\theta} ... Quotient\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] \cos^2\theta\tan\theta + 1 \\[3ex] = \cos^2\theta * \dfrac{\sin\theta}{\cos\theta} + 1 \\[5ex] = \cos\theta\sin\theta + 1 \\[3ex] \underline{Denominator} \\[3ex] \cos^2\theta\tan\theta - 1 \\[3ex] = \cos^2\theta * \dfrac{\sin\theta}{\cos\theta} - 1 \\[5ex] = \cos\theta\sin\theta - 1 \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{\cos\theta\sin\theta + 1}{\cos\theta\sin\theta - 1} ... Same\:\: Simplified\:\: Result$
(31.) $\sec\omega + \tan\omega = \dfrac{\cos\omega}{1 - \sin\omega}$

$\underline{LHS} \\[3ex] \sec\omega + \tan\omega \\[3ex] \sec\omega = \dfrac{1}{\cos\omega} ... Reciprocal\:\: Identity \\[5ex] \tan\omega = \dfrac{\sin\omega}{\cos\omega} ... Quotient\:\: Identity \\[5ex] = \dfrac{1}{\cos\omega} + \dfrac{\sin\omega}{\cos\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\cos\omega} \\[5ex]$ The $LHS$ is simplified
The $RHS$ is also simplified
But, they are not the same
So, we have to do something ...
Let us multiply the numerator and the denominator of the $LHS$ by $1 - \sin\omega$

Student: Why do we have to multiply it by $1 - \sin\omega$?
Why not multiply it with something else besides $1 - \sin\omega$?
Teacher: Good question
We can multiply it by anything provided we get the same result as the $RHS$
However, I suggest $1 - \sin\omega$ because of the "Difference of Two Squares" as regards the numerator on the $LHS$
Let us try it with that
If it does not give us the $RHS$, then we can find another term to multiply with.

$= \dfrac{1 + \sin\omega}{\cos\omega} * \dfrac{1 - \sin\omega}{1 - \sin\omega} \\[5ex] = \dfrac{(1 + \sin\omega)(1 - \sin\omega)}{\cos\omega(1 - \sin\omega)} \\[5ex] (1 + \sin\omega)(1 - \sin\omega) = 1^2 - \sin^2\omega = 1 - \sin^2\omega ... Difference \:\:of\:\: Two\:\: Squares \\[3ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \cos^2\omega = 1 - \sin^2\omega \\[3ex] = \dfrac{\cos^2\omega}{\cos\omega(1 - \sin\omega)} \\[5ex] = \dfrac{\cos\omega}{1 - \sin\omega} \\[5ex] = RHS$
(32.) $2\csc^2 \theta - 1 = \dfrac{1 + \cos^2 \theta}{\sin^2 \theta}$

$\underline{RHS} \\[3ex] \dfrac{1 + \cos^2 \theta}{\sin^2 \theta} \\[3ex] \dfrac{1}{\sin^2 \theta} + \dfrac{\cos^2 \theta}{\sin^2 \theta} \\[5ex] \dfrac{1}{\sin^2 \theta} = \csc^2 \theta ... Reciprocal\:\: Identity \\[5ex] \sin^2 \theta + \cos^2 \theta = 1 ...Pythagorean\:\: Identity \\[3ex] \cos^2 \theta = 1 - \sin^2 \theta \\[3ex] \rightarrow \csc^2 \theta + \dfrac{1 - \sin^2 \theta}{\sin^2 \theta} \\[5ex] = \csc^2 \theta + \dfrac{1}{\sin^2 \theta} - \dfrac{\sin^2 \theta}{\sin^2 \theta} \\[5ex] = \csc^2 \theta + \csc^2 \theta - 1 \\[3ex] = 2\csc^2 \theta - 1 \\[3ex] = LHS$
(33.) $1 - \cos5\alpha\cos3\alpha - \sin5\alpha\sin3\alpha = 2\sin^2\alpha$

$\underline{LHS} \\[3ex] 1 - \cos5\alpha\cos3\alpha - \sin5\alpha\sin3\alpha \\[3ex] = 1 - (\cos5\alpha\cos3\alpha + \sin5\alpha\sin3\alpha) \\[3ex] \cos5\alpha\cos3\alpha + \sin5\alpha\sin3\alpha = \cos(5\alpha - 3\alpha) ... Difference\:\: Formula \\[3ex] = 1 - \cos(5\alpha - 3\alpha) \\[3ex] = 1 - \cos2\alpha \\[3ex] \cos2\alpha = \cos^2\alpha - \sin^2\alpha ... Double-Angle\:\: Formula \\[3ex] = 1 - (\cos^2\alpha - \sin^2\alpha) \\[3ex] = 1 - \cos^2\alpha + \sin^2\alpha \\[3ex] \sin^2\alpha + \cos^2\alpha = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \cos^2\alpha = 1 - \sin^2\alpha \\[3ex] = 1 - (1 - \sin^2\alpha) + \sin^2\alpha \\[3ex] = 1 - 1 + \sin^2\alpha + \sin^2\alpha \\[3ex] = 2\sin^2\alpha \\[3ex] = RHS$
(34.) $\dfrac{1}{\cos^2\beta - \sin^2\beta} = \dfrac{\cot\beta + \tan\beta}{\cot\beta - \tan\beta}$

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