Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right) = \dfrac{\sin \beta}{\sin \alpha}$

LHS $= \dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right)$

$= \dfrac{\sin \left[\left(\alpha + \alpha\right) + \beta\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha }$

$= \dfrac{\sin \left[\alpha + \left(\alpha + \beta\right)\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \alpha \cos \left(\alpha + \beta\right) + \cos \alpha \sin \left(\alpha + \beta\right) - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \left(\alpha + \beta\right) \cos \alpha - \cos \left(\alpha + \beta\right) \sin \alpha}{\sin \alpha}$

$= \dfrac{\sin \left[\left(\alpha + \beta\right) - \alpha\right]}{\sin \alpha}$

$= \dfrac{\sin \beta}{\sin \alpha} = $ RHS

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right) = 2 \sec 2 \alpha$

LHS $=\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right)$

$= \dfrac{1}{\cos \left(\dfrac{\pi}{4} + \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}\right) + \alpha \right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\dfrac{\pi}{2} - \left(\dfrac{\pi}{4} - \alpha\right)\right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{2 \sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{\sin \left[2 \left(\dfrac{\pi}{4} - \alpha\right)\right]}$

$= \dfrac{2}{\sin \left(\dfrac{\pi}{2} - 2 \alpha\right)}$

$= \dfrac{2}{\cos 2 \alpha}$

$= 2 \sec 2 \alpha = $ RHS

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\cot \alpha + \cot \left(270^\circ + \alpha\right)}{\cot \alpha - \cot \left(270^\circ + \alpha\right)} + 2 \cos \left(135^\circ + \alpha\right) \cos \left(135^\circ - \alpha\right) = 2 \cos 2 \alpha$

$\begin{aligned} LHS & = \dfrac{\cot \alpha + \cot \left(270^\circ + \alpha\right)}{\cot \alpha - \cot \left(270^\circ + \alpha\right)} + 2 \cos \left(135^\circ + \alpha\right) \cos \left(135^\circ - \alpha\right) \\\\ & = \dfrac{\cot \alpha - \tan \alpha}{\cot \alpha + \tan \alpha} + 2 \cos \left(\dfrac{270^\circ + 2 \alpha}{2}\right) \cos \left(\dfrac{270^\circ - 2 \alpha}{2}\right) \\\\ & = \dfrac{\cot \alpha - \tan \alpha}{\cot \alpha + \tan \alpha} + \left[\cos \left(270^\circ\right) + \cos 2 \alpha\right] \\\\ & = \dfrac{\dfrac{1}{\tan \alpha} - \tan \alpha}{\dfrac{1}{\tan \alpha} + \tan \alpha} + 0 + \cos 2 \alpha \\\\ & = \dfrac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} + \cos 2 \alpha \\\\ & = \dfrac{1 - \dfrac{\sin^2 \alpha}{\cos^2 \alpha}}{1 + \dfrac{\sin^2 \alpha}{\cos^2 \alpha}} + \cos 2 \alpha \\\\ & = \dfrac{\cos^2 \alpha - \sin^2 \alpha}{\cos^2 \alpha + \sin^2 \alpha} + \cos 2 \alpha \\\\ & = \dfrac{\cos 2 \alpha}{1} + \cos 2 \alpha \\\\ & = 2 \cos 2 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $2 \left(\sin^6 \alpha + \cos^6 \alpha\right) - 3 \left(\sin^4 \alpha + \cos^4 \alpha\right) + 1 = 0$

$\begin{aligned} LHS & = 2 \left(\sin^6 \alpha + \cos^6 \alpha\right) - 3 \left(\sin^4 \alpha + \cos^4 \alpha\right) + 1 \\\\ & = 2 \left[\left(\sin^2 \alpha\right)^3 + \left(\cos^2 \alpha\right)^3\right] - 3 \sin^4 \alpha - 3 \cos^4 \alpha + 1 \\\\ & = 2 \left[\left(\sin^2 \alpha + \cos^2 \alpha\right) \left(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha\right)\right] \\ & \hspace{6cm} - 3 \sin^4 \alpha - 3 \cos^4 \alpha + 1 \\\\ & = 2 \sin^4 \alpha - 2 \sin^2 \alpha \cos^2 \alpha + 2 \cos^4 \alpha - 3 \sin^4 \alpha - 3 \cos^4 \alpha + 1 \\\\ & = - \sin^4 \alpha - 2 \sin^2 \alpha \cos^2 \alpha - \cos^4 \alpha + 1 \\\\ & = - \left[\sin^4 \alpha + 2 \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha\right] + 1 \\\\ & = - \left[\sin^2 \alpha + \cos^2 \alpha\right]^2 + 1 \\\\ & = - 1 + 1 \\\\ & = 0 = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\cos^6 \alpha - \sin^6 \alpha = \dfrac{\left(3 + \cos^2 2 \alpha\right) \cos 2 \alpha}{4}$

$\begin{aligned} LHS & = \cos^6 \alpha - \sin^6 \alpha \\\\ & = \left(\cos^2 \alpha\right)^3 - \left(\sin^2 \alpha\right)^3 \\\\ & = \left[\cos^2 \alpha - \sin^2 \alpha\right] \left[\cos^4 \alpha + \cos^2 \alpha \sin^2 \alpha + \sin^4 \alpha\right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha \left(\cos^2 \alpha + \sin^2 \alpha\right) + \sin^4 \alpha\right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha + \sin^4 \alpha\right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha + \left(\sin^2 \alpha\right)^2 \right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha + \left(\dfrac{1 - \cos 2 \alpha}{2}\right)^2\right] \\\\ & = \cos 2 \alpha \times \left[\cos^2 \alpha + \dfrac{1 - 2 \cos 2 \alpha + \cos^2 2 \alpha}{4}\right] \\\\ & = \cos 2 \alpha \times \left[\dfrac{4 \cos^2 \alpha + 1 - 2 \left(2 \cos^2 \alpha - 1\right) + \cos^2 2 \alpha}{4}\right] \\\\ & = \cos 2 \alpha \times \left[\dfrac{4 \cos^2 \alpha + 1 - 4 \cos^2 \alpha + 2 + \cos^2 2 \alpha}{4}\right] \\\\ & = \cos 2 \alpha \times \left[\dfrac{3 + \cos^2 2 \alpha}{4}\right] = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{3 - 4 \cos 2 \alpha + \cos 4 \alpha}{3 + 4 \cos 2 \alpha + \cos 4 \alpha} = {\tan^4 \alpha}$

$\begin{aligned} LHS & = \dfrac{3 - 4 \cos 2 \alpha + \cos 4 \alpha}{3 + 4 \cos 2 \alpha + \cos 4 \alpha} \\\\ & = \dfrac{3 - 4 \cos 2 \alpha + 1 - 2 \sin^2 2 \alpha}{3 + 4 \cos 2 \alpha + 1 - 2 \sin^2 2 \alpha} \\\\ & = \dfrac{4 - 4 \cos 2 \alpha - 2 \sin^2 2 \alpha}{4 + 4 \cos 2 \alpha - 2 \sin^2 2 \alpha} \\\\ & = \dfrac{2 \left(1 - \cos 2 \alpha\right) - \sin^2 2 \alpha}{2 \left(1 + \cos 2 \alpha \right) - \sin^2 2 \alpha} \\\\ & = \dfrac{2 \times 2 \sin^2 \alpha - \left(2 \sin \alpha \cos \alpha\right)^2}{2 \times 2 \cos^2 \alpha - \left(2 \sin \alpha \cos \alpha\right)^2} \\\\ & = \dfrac{4 \sin^2 \alpha - 4 \sin^2 \alpha \cos^2 \alpha}{4 \cos^2 \alpha - 4 \sin^2 \alpha \cos^2 \alpha} \\\\ & = \dfrac{\sin^2 \alpha - \sin^2 \alpha \cos^2 \alpha}{\cos^2 \alpha - \sin^2 \alpha \cos^2 \alpha} \\\\ & = \dfrac{\sin^2 \alpha \left(1 - \cos^2 \alpha\right)}{\cos^2 \alpha \left(1 - \sin^2 \alpha\right)} \\\\ & = \dfrac{\sin^2 \alpha \times \sin^2 \alpha}{\cos^2 \alpha \times \cos^2 \alpha} \\\\ & = \dfrac{\sin^4 \alpha}{\cos^4 \alpha} \\\\ & = \tan^4 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\cos 2 \alpha - \cos 3 \alpha - \cos 4 \alpha + \cos 5 \alpha = - 4 \sin \left(\dfrac{\alpha}{2}\right) \sin \alpha \cos \left(\dfrac{7 \alpha}{2}\right)$

$\begin{aligned} LHS & = \cos 2 \alpha - \cos 3 \alpha - \cos 4 \alpha + \cos 5 \alpha \\\\ & = \left(\cos 5 \alpha + \cos 2 \alpha\right) - \left(\cos 4 \alpha + \cos 3 \alpha\right) \\\\ & = 2 \cos \left(\dfrac{5 \alpha + 2 \alpha}{2}\right) \cos \left(\dfrac{5 \alpha - 2 \alpha}{2}\right) - 2 \cos \left(\dfrac{4 \alpha + 3 \alpha}{2}\right) \cos \left(\dfrac{4 \alpha - 3 \alpha}{2}\right) \\\\ & = 2 \cos \left(\dfrac{7 \alpha}{2}\right) \cos \left(\dfrac{3 \alpha}{2}\right) - 2 \cos \left(\dfrac{7 \alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right) \\\\ & = 2 \cos \left(\dfrac{7 \alpha}{2}\right) \left[\cos \left(\dfrac{3 \alpha}{2}\right) - \cos \left(\dfrac{\alpha}{2}\right)\right] \\\\ & = 2 \cos \left(\dfrac{7 \alpha}{2}\right) \left[- 2 \sin \left(\dfrac{\dfrac{3 \alpha}{2} - \dfrac{\alpha}{2}}{2}\right) \sin \left(\dfrac{\dfrac{3 \alpha}{2} + \dfrac{\alpha}{2}}{2}\right) \right] \\\\ & = - 4 \cos \left(\dfrac{7 \alpha}{2}\right) \sin \left(\dfrac{\alpha}{2}\right) \sin \left(\dfrac{2 \alpha}{2}\right) \\\\ & = - 4 \sin \left(\dfrac{\alpha}{2}\right) \sin \alpha \cos \left(\dfrac{7 \alpha}{2}\right) = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\sin x + \cos x}{\sin x - \cos x} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} = \dfrac{2}{\tan x + 1}$

$\begin{aligned} LHS & = \dfrac{\sin x + \cos x}{\sin x - \cos x} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\dfrac{\sin x}{\cos x} + 1}{\dfrac{\sin x}{\cos x} - 1} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\tan x + 1}{\tan x - 1} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\left(\tan x + 1\right)^2}{\left(\tan x - 1\right) \left(\tan x + 1\right)} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\tan^2 x + 1 + 2 \tan x}{\tan^2 x - 1} - \dfrac{\sec^2 x + 2}{\tan^2 x - 1} \\\\ & = \dfrac{\sec^2 x + 2 \tan x - sec^2 x - 2}{\tan^2 x - 1} \\\\ & = \dfrac{2 \left(\tan x - 1\right)}{\left(\tan x + 1\right) \left(\tan x - 1\right)} \\\\ & = \dfrac{2}{\tan x + 1} = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\sqrt{1 + \cos \alpha} + \sqrt{1 - \cos \alpha}}{\sqrt{1 + \cos \alpha} - \sqrt{1 - \cos \alpha}} = \cot \left(\dfrac{\alpha}{2} + \dfrac{\pi}{4}\right)$, $\;\;\;$ $\pi < \alpha < 2 \pi$

$\begin{aligned} LHS & = \dfrac{\sqrt{1 + \cos \alpha} + \sqrt{1 - \cos \alpha}}{\sqrt{1 + \cos \alpha} - \sqrt{1 - \cos \alpha}} \\\\ & = \dfrac{\left(\sqrt{1 + \cos \alpha} + \sqrt{1 - \cos \alpha}\right)^2}{\left(\sqrt{1 + \cos \alpha} - \sqrt{1 - \cos \alpha}\right) \left(\sqrt{1 + \cos \alpha} + \sqrt{1 - \cos \alpha}\right)} \\\\ & = \dfrac{1 + \cos \alpha + 1 - \cos \alpha - 2 \sqrt{\left(1 + \cos \alpha\right) \left(1 - \cos \alpha\right)}}{1 + \cos \alpha - \left(1 - \cos \alpha\right)} \\\\ & = \dfrac{2 - 2 \sqrt{1 - \cos^2 \alpha}}{2 \cos \alpha} \\\\ & = \dfrac{1 - \sqrt{\sin^2 \alpha}}{\cos \alpha} \\\\ & = \dfrac{1 - \sin \alpha}{\cos \alpha} \\\\ & = \dfrac{\sin^2 \left(\dfrac{\alpha}{2}\right) + \cos^2 \left(\dfrac{\alpha}{2}\right) - 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right)}{\cos^2 \left(\dfrac{\alpha}{2}\right) - \sin^2 \left(\dfrac{\alpha}{2}\right)} \\\\ & = \dfrac{\left[\cos \left(\dfrac{\alpha}{2}\right) - \sin \left(\dfrac{\alpha}{2}\right)\right]^2}{\left[\cos \left(\dfrac{\alpha}{2}\right) + \sin \left(\dfrac{\alpha}{2}\right)\right] \left[\cos \left(\dfrac{\alpha}{2}\right) - \sin \left(\dfrac{\alpha}{2}\right)\right]} \\\\ & = \dfrac{\cos \left(\dfrac{\alpha}{2}\right) - \sin \left(\dfrac{\alpha}{2}\right)}{\cos \left(\dfrac{\alpha}{2}\right) + \sin \left(\dfrac{\alpha}{2}\right)} \\\\ & = \dfrac{1 - \dfrac{\sin \left(\dfrac{\alpha}{2}\right)}{\cos \left(\dfrac{\alpha}{2}\right)}}{1 + \dfrac{\sin \left(\dfrac{\alpha}{2}\right)}{\cos \left(\dfrac{\alpha}{2}\right)}} \\\\ & = \dfrac{1 - \tan \left(\dfrac{\alpha}{2}\right)}{1 + \tan \left(\dfrac{\alpha}{2}\right)} \\\\ & = \dfrac{1 - \tan \left(\dfrac{\alpha}{2}\right) \tan \left(\dfrac{\pi}{4}\right)}{\tan \left(\dfrac{\alpha}{2}\right) + \tan \left(\dfrac{\pi}{4}\right)} \\\\ & = \cot \left(\dfrac{\alpha}{2} + \dfrac{\pi}{4}\right) = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\sin^2 2 \alpha + 4 \sin^4 \alpha - 4 \sin^2 \alpha \cos^2 \alpha}{4 - \sin^2 2 \alpha - 4 \sin^2 \alpha} = \tan^4 \alpha$

$\begin{aligned} LHS & = \dfrac{\sin^2 2 \alpha + 4 \sin^4 \alpha - 4 \sin^2 \alpha \cos^2 \alpha}{4 - \sin^2 2 \alpha - 4 \sin^2 \alpha} \\\\ & = \dfrac{\left(2 \sin \alpha \cos \alpha\right)^2 - 4 \sin^2 \alpha \cos^2 \alpha + 4 \sin^4 \alpha}{4 - \left(2 \sin \alpha \cos \alpha\right)^2 - 4 \sin^2 \alpha} \\\\ & = \dfrac{4 \sin^2 \alpha \cos^2 \alpha - 4 \sin^2 \alpha \cos^2 \alpha + 4 \sin^4 \alpha}{4 - 4 \sin^2 \alpha \cos^2 \alpha - 4 \sin^2 \alpha} \\\\ & = \dfrac{\sin^4 \alpha}{\left(1 - \sin^2 \alpha\right) - \sin^2 \alpha \cos^2 \alpha} \\\\ & = \dfrac{\sin^4 \alpha}{\cos^2 \alpha - \sin^2 \alpha \cos^2 \alpha} \\\\ & = \dfrac{\sin^4 \alpha}{\cos^2 \alpha \left(1 - \sin^2 \alpha\right)} \\\\ & = \dfrac{\sin^4 \alpha}{\cos^2 \alpha \times \cos^2 \alpha} \\\\ & = \dfrac{\sin^4 \alpha}{\cos^4 \alpha} \\\\ & = \tan^4 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $3 - 4 \cos 2 \alpha + \cos 4 \alpha = 8 \sin^4 \alpha$

$\begin{aligned} LHS & = 3 - 4 \cos 2 \alpha + \cos 4 \alpha \\\\ & = 3 - 4 \left(1 - 2 \sin^2 \alpha\right) + \cos \left[2 \left(2 \alpha\right)\right] \\\\ & = 3 - 4 + 8 \sin^2 \alpha + 1 - 2 \sin^2 2 \alpha \\\\ & = 8 \sin^2 \alpha - 2 \sin^2 2 \alpha \\\\ & = 8 \sin^2 \alpha - 2 \left(2 \sin \alpha \cos \alpha\right)^2 \\\\ & = 8 \sin^2 \alpha - 2 \times 4 \sin^2 \alpha \cos^2 \alpha \\\\ & = 8 \sin^2 \alpha \left(1 - \cos^2 \alpha\right) \\\\ & = 8 \sin^2 \alpha \times \sin^2 \alpha \\\\ & = 8 \sin^4 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{1 - \cos 2 \alpha + \sin 2 \alpha}{1 + \cos 2 \alpha + \sin 2 \alpha} = \tan \alpha$

$\begin{aligned} LHS & = \dfrac{1 - \cos 2 \alpha + \sin 2 \alpha}{1 + \cos 2 \alpha + \sin 2 \alpha} \\\\ & = \dfrac{\left(1 - \cos 2 \alpha\right) + 2 \sin \alpha \cos \alpha}{\left(1 + \cos 2 \alpha\right) + 2 \sin \alpha \cos \alpha} \\\\ & = \dfrac{2 \sin^2 \alpha + 2 \sin \alpha \cos \alpha}{2 \cos^2 \alpha + 2 \sin \alpha \cos \alpha} \\\\ & = \dfrac{2 \sin \alpha \left(\sin \alpha + \cos \alpha\right)}{2 \cos \alpha \left(\cos \alpha + \sin \alpha\right)} \\\\ & = \dfrac{\sin \alpha}{\cos \alpha} \\\\ & = \tan \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\cot \alpha - \tan \alpha - 2 \tan 2 \alpha - 4 \tan 4 \alpha = 8 \cot 8 \alpha$

$\begin{aligned} LHS & = \cot \alpha - \tan \alpha - 2 \tan 2 \alpha - 4 \tan 4 \alpha \\\\ & = \dfrac{\cos \alpha}{\sin \alpha} - \dfrac{\sin \alpha}{\cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{\cos^2 \alpha - \sin^2 \alpha}{\sin \alpha \cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{\cos 2 \alpha}{\sin \alpha \cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{2 \cos 2 \alpha}{2 \sin \alpha \cos \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{2 \cos 2 \alpha}{\sin 2 \alpha} - \dfrac{2 \sin 2 \alpha}{\cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{2 \left(\cos^2 2 \alpha - \sin^2 2 \alpha\right)}{\sin 2 \alpha \cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{2 \cos 4 \alpha}{\sin 2 \alpha \cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{4 \cos 4 \alpha}{2 \sin 2 \alpha \cos 2 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{4 \cos 4 \alpha}{\sin 4 \alpha} - \dfrac{4 \sin 4 \alpha}{\cos 4 \alpha} \\\\ & = \dfrac{4 \left(\cos^2 4 \alpha - \sin^2 4 \alpha\right)}{\sin 4 \alpha \cos 4 \alpha} \\\\ & = \dfrac{4 \cos 8 \alpha}{\sin 4 \alpha \cos 4 \alpha} \\\\ & = \dfrac{8 \cos 8 \alpha}{2 \sin 4 \alpha \cos 4 \alpha} \\\\ & = \dfrac{8 \cos 8 \alpha}{\sin 8 \alpha} \\\\ & = 8 \cot 8 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\left(\sin \alpha - \sin \beta\right) \left(\sin \alpha + \sin \beta\right) = \sin \left(\alpha - \beta\right) \sin \left(\alpha + \beta\right)$

$\begin{aligned} RHS & = \sin \left(\alpha - \beta\right) \sin \left(\alpha + \beta\right) \\\\ & = \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right) \left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right) \\\\ & = \sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta \\\\ & = \sin^2 \alpha \left(1 - \sin^2 \beta\right) - \sin^2 \beta \left(1 - \sin^2 \alpha\right) \\\\ & = \sin^2 \alpha - \sin^2 \alpha \sin^2 \beta - \sin^2 \beta + \sin^2 \alpha \sin^2 \beta \\\\ & = \sin^2 \alpha - \sin^2 \beta \\\\ & = \left(\sin \alpha - \sin \beta\right) \left(\sin \alpha + \sin \beta\right) = LHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\tan 3 \alpha}{\tan \alpha} = \dfrac{3 - \tan^2 \alpha}{1 - 3 \tan^2 \alpha}$

$\begin{aligned} LHS & = \dfrac{\tan 3 \alpha}{\tan \alpha} \\\\ & = \dfrac{\tan \left(2 \alpha + \alpha\right)}{\tan \alpha} \\\\ & = \left(\dfrac{\tan 2 \alpha + \tan \alpha}{1 - \tan 2 \alpha \; \tan \alpha}\right) \times \dfrac{1}{\tan \alpha} \\\\ & = \left(\dfrac{\dfrac{2 \tan \alpha}{1 - \tan^2 \alpha} + \tan \alpha}{1 - \dfrac{2 \tan \alpha}{1 - \tan^2 \alpha}\times \tan \alpha}\right) \times \dfrac{1}{\tan \alpha} \\\\ & = \left(\dfrac{2 \tan \alpha + \tan \alpha - \tan^3 \alpha}{1 - \tan^2 \alpha - 2 \tan^2 \alpha}\right) \times \dfrac{1}{\tan \alpha} \\\\ & = \left(\dfrac{3 \tan \alpha - \tan^3 \alpha}{1 - 3 \tan^2 \alpha}\right) \times \dfrac{1}{\tan \alpha} \\\\ & = \dfrac{3 - \tan^2 \alpha}{1 - 3 \tan^2 \alpha} = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\cot \alpha + \tan \alpha}{1 + \tan 2 \alpha \tan \alpha} = 2 \cot 2 \alpha$

$\begin{aligned} \cot \alpha + \tan \alpha & = \dfrac{\cos \alpha}{\sin \alpha} + \dfrac{\sin \alpha}{\cos \alpha} \\\\ & = \dfrac{\cos^2 \alpha + \sin^2 \alpha}{\sin \alpha \cos \alpha} \\\\ & = \dfrac{1}{\sin \alpha \cos \alpha} \\\\ & = \dfrac{2}{2 \sin \alpha \cos \alpha} \\\\ & = \dfrac{2}{\sin 2 \alpha} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} 1 + \tan 2 \alpha \tan \alpha & = 1 + \left(\dfrac{2 \tan \alpha}{1 - \tan^2 \alpha}\right) \tan \alpha \\\\ & = \dfrac{1 - \tan^2 \alpha + 2 \tan^2 \alpha}{1 - \tan^2 \alpha} \\\\ & = \dfrac{1 + \tan^2 \alpha}{1 - \tan^2 \alpha} \\\\ & = \dfrac{1 + \dfrac{\sin^2 \alpha}{\cos^2 \alpha}}{1 - \dfrac{\sin^2 \alpha}{\cos^2 \alpha}} \\\\ & = \dfrac{\cos^2 \alpha + \sin^2 \alpha}{\cos^2 \alpha - \sin^2 \alpha} \\\\ & = \dfrac{1}{\cos^2 \alpha - \sin^2 \alpha} \\\\ & = \dfrac{1}{\cos 2 \alpha} \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \dfrac{\cot \alpha + \tan \alpha}{1 + \tan 2 \alpha \tan \alpha} \\\\ & = \dfrac{\dfrac{2}{\sin 2 \alpha}}{\dfrac{1}{\cos 2 \alpha}} \;\;\; \left[\text{By equations (1) and (2)}\right] \\\\ & = 2 \times \dfrac{\cos 2 \alpha}{\sin 2 \alpha} \\\\ & = 2 \cot 2 \alpha = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{1 - 2 \cos^2 \alpha}{2 \tan \left(\alpha - \dfrac{\pi}{4}\right) \sin^2 \left(\dfrac{\pi}{4} + \alpha\right)} = 1$

$\begin{aligned} \tan \left(\alpha - \dfrac{\pi}{4}\right) & = \dfrac{\tan \alpha - \tan \left(\dfrac{\pi}{4}\right)}{1 + \tan \alpha \tan \left(\dfrac{\pi}{4}\right)} \\\\ & = \dfrac{\tan \alpha - 1}{1 + \tan \alpha} \\\\ & = \dfrac{\dfrac{\sin \alpha}{\cos \alpha} - 1}{\dfrac{\sin \alpha}{\cos \alpha} + 1} \\\\ & = \dfrac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \sin \left(\dfrac{\pi}{4} + \alpha\right) & = \sin \left(\dfrac{\pi}{4}\right) \cos \alpha + \cos \left(\dfrac{\pi}{4}\right) \sin \alpha \\\\ & = \dfrac{1}{\sqrt{2}} \times \cos \alpha + \dfrac{1}{\sqrt{2}} \times \sin \alpha \\\\ & = \dfrac{1}{\sqrt{2}} \left(\sin \alpha + \cos \alpha\right) \end{aligned}$

$\begin{aligned} \therefore \; \sin^2 \left(\dfrac{\pi}{4} + \alpha\right) & = \dfrac{1}{2} \left(\sin \alpha + \cos \alpha\right) ^2 \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \dfrac{1 - 2 \cos^2 \alpha}{2 \tan \left(\alpha - \dfrac{\pi}{4}\right) \sin^2 \left(\dfrac{\pi}{4} + \alpha\right)} \\\\ & = \dfrac{- \left(2 \cos^2 \alpha - 1\right)}{2 \left(\dfrac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}\right) \times \dfrac{1}{2} \left(\sin \alpha + \cos \alpha\right)^2} \\ & \hspace{2cm} \;\;\; \left[\text{by equations (1) and (2)}\right] \\\\ & = \dfrac{- \cos 2 \alpha}{- \left(\cos \alpha - \sin \alpha\right) \left(\sin \alpha + \cos \alpha\right)} \\\\ & = \dfrac{- \cos 2 \alpha}{- \left(\cos^2 \alpha - \sin^2 \alpha\right)} \\\\ & = \dfrac{- \cos 2 \alpha}{- \cos 2 \alpha} \\\\ & = 1 = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{\tan \left(x - \dfrac{\pi}{2}\right) \cos \left(\dfrac{3 \pi}{2} + x\right) - \sin^3 \left(\dfrac{7 \pi}{2} - x\right)}{\cos \left(x - \dfrac{\pi}{2}\right) \tan \left(\dfrac{3 \pi}{2} + x\right)} = \sin^2 x$

$\begin{aligned} \tan \left(x - \dfrac{\pi}{2}\right) & = \tan \left[- \left(\dfrac{\pi}{2} - x\right)\right] \\\\ & = - \tan \left(\dfrac{\pi}{2} - x\right) \\\\ & = - \cot x \\\\ & = \dfrac{- \cos x}{\sin x} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \cos \left(\dfrac{3 \pi}{2} + x\right) & = \cos \left[\pi + \left(\dfrac{\pi}{2} + x\right)\right] \\\\ & = -\cos \left(\dfrac{\pi}{2} + x\right) \\\\ & = - \left(- \sin x\right) \\\\ & = \sin x \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} \sin \left(\dfrac{7 \pi}{2} - x\right) & = \sin \left[\dfrac{\pi}{2} + \left(3 \pi - x\right)\right] \\\\ & = \cos \left(3 \pi - x\right) \\\\ & = - \cos x \;\;\; \cdots \; (3) \end{aligned}$

$\begin{aligned} \cos \left(x - \dfrac{\pi}{2}\right) & = \cos \left[- \left(\dfrac{\pi}{2} - x\right)\right] \\\\ & = \cos \left(\dfrac{\pi}{2} - x\right) \\\\ & = \sin x \;\;\; \cdots \; (4) \end{aligned}$

$\begin{aligned} \tan \left(\dfrac{3 \pi}{2} + x\right) & = \tan \left[\pi + \left(\dfrac{\pi}{2} + x\right)\right] \\\\ & = \tan \left(\dfrac{\pi}{2} + x\right) \\\\ & = \dfrac{- \cos x}{\sin x} \;\;\; \cdots \; (5) \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \dfrac{\tan \left(x - \dfrac{\pi}{2}\right) \cos \left(\dfrac{3 \pi}{2} + x\right) - \sin^3 \left(\dfrac{7 \pi}{2} - x\right)}{\cos \left(x - \dfrac{\pi}{2}\right) \tan \left(\dfrac{3 \pi}{2} + x\right)} \\\\ & = \dfrac{\dfrac{- \cos x}{\sin x} \times \sin x - \left(- \cos x\right)^3}{\sin x \times \left(\dfrac{- \cos x}{\sin x}\right)} \;\;\; \left[\text{In view of equations } (1) \cdots (5) \right] \\\\ & = \dfrac{- \cos x + \cos^3 x}{- \cos x} \\\\ & = 1 - \cos^2 x \\\\ & = \sin^2 x = RHS \end{aligned}$

Hence proved.

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{1 - \sin 2 \alpha}{1 + \sin 2 \alpha} = \cot^2 \left(\dfrac{\pi}{4} + \alpha\right)$

$\begin{aligned} \cot \left(\dfrac{\pi}{4} + \alpha\right) & = \dfrac{1}{\tan \left(\dfrac{\pi}{4} + \alpha\right)} \\\\ & = \dfrac{1 - \tan \left(\dfrac{\pi}{4}\right) \tan \alpha}{\tan \dfrac{\pi}{4} + \tan \alpha} \\\\ & = \dfrac{1 - \tan \alpha}{1 + \tan \alpha} \\\\ & = \dfrac{1 - \dfrac{\sin \alpha}{\cos \alpha}}{1 + \dfrac{\sin \alpha}{\cos \alpha}} \\\\ & = \dfrac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha} \end{aligned}$

$\begin{aligned} \therefore \; RHS & = \cot^2 \left(\dfrac{\pi}{4} + \alpha\right) \\\\ & = \left(\dfrac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}\right)^2 \\\\ & = \dfrac{\cos^2 \alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha}{\cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha} \\\\ & = \dfrac{1 - \sin 2 \alpha}{1 + \sin 2 \alpha} = LHS \end{aligned}$

Hence proved.

Identity Transformation of Trigonometric Expressions

Prove the identity: $\dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha } - \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} = 1$

To Prove That (TPT) $\;$ $\dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha } - \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} = 1$

i.e. $\;$ TPT $\;$ $1 + \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} = \dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha}$

$\begin{aligned} LHS & = 1 + \dfrac{\left(1 - \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} \\\\ & = \dfrac{4 \tan^2 \alpha + 1 + \tan^4 \alpha - 2 \tan^2 \alpha}{4 \tan^2 \alpha} \\\\ & = \dfrac{1 + \tan^4 \alpha + 2 \tan^2 \alpha}{4 \tan^2 \alpha} \\\\ & = \dfrac{\left(1 + \tan^2 \alpha\right)^2}{4 \tan^2 \alpha} \\\\ & = \dfrac{\left(\sec^2 \alpha\right)^2}{4 \tan^2 \alpha} \\\\ & = \dfrac{1}{\cos^4 \alpha} \times \dfrac{\cos^2 \alpha}{4 \sin^2 \alpha} \\\\ & = \dfrac{1}{4 \sin^2 \alpha \cos^2 \alpha} = RHS \end{aligned}$

Hence proved.