Trigonometry - Trigonometric Functions

If $\;$ $\text{cosec} \theta - \sin \theta = a^3$, $\;$ $\sec \theta - \cos \theta = b^3$,
then prove that $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$

Given: $\;$ $\text{cosec } \theta - \sin \theta = a^3$

i.e. $\;$ $\dfrac{1}{\sin \theta} - \sin \theta = a^3$

i.e. $\;$ $a^3 = \dfrac{1 - \sin^2 \theta}{\sin \theta}$

i.e. $\;$ $a^3 = \dfrac{\cos^2 \theta}{\sin \theta}$ $\;\;\; \cdots \; (1)$

and $\;$ $\sec \theta - \cos \theta = b^3$

i.e. $\;$ $\dfrac{1}{\cos \theta} - \cos \theta = b^3$

i.e. $\;$ $b^3 = \dfrac{1 - \cos^2 \theta}{\cos \theta}$

i.e. $\;$ $b^3 = \dfrac{\sin^2 \theta}{\cos \theta}$ $\;\;\; \cdots \; (2)$

Dividing equations $(1)$ and $(2)$ gives

$\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \div \dfrac{\cos^2 \theta}{\sin \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \times \dfrac{\sin \theta}{\cos^2 \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^3 \theta}{\cos^3 \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \tan^3 \theta$

i.e. $\;$ $\dfrac{b}{a} = \tan \theta$

$\implies$ $\sin \theta = \dfrac{b}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (3)$

and $\;$ $\cos \theta = \dfrac{a}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (4)$

Substituting the values of $\sin \theta$ and $\cos \theta$ from equations $(3)$ and $(4)$ in equation $(1)$ gives

$a^3 = \dfrac{a^2}{a^2 + b^2} \div \dfrac{b}{\sqrt{a^2 + b^2}}$

i.e. $\;$ $a^3 = \dfrac{a^2}{a^2 + b^2} \times \dfrac{\sqrt{a^2 + b^2}}{b}$

i.e. $\;$ $a \; b = \dfrac{1}{\sqrt{a^2 + b^2}}$

i.e. $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$

Hence proved

Trigonometric Functions

Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$

$\cot \left(\alpha - \beta\right) = \dfrac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}$

$\implies$ $\cot \beta - \cot \alpha = \dfrac{\cot \alpha \cot \beta + 1}{\cot \left(\alpha - \beta\right)}$

$\begin{aligned} \text{LHS} & = \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x \\ & \\ & = \cot 2x \left(\cot x - \cot 3x\right) - \cot 3x \cot x \\ & \\ & = \cot 2x \left[\dfrac{\cot x \cot 3x + 1}{\cot \left(3x-x\right)}\right] - \cot 3x \cot x \\ & \\ & = \dfrac{\cot 2x \left(\cot x \cot 3x + 1\right)}{\cot 2x} - \cot 3x \cot x \\ & \\ & = \cot 3x \cot x + 1 - \cot 3x \cot x \\ & \\ & = 1 = \text{RHS} \end{aligned}$

Trigonometric Functions

Prove that
$\dfrac{\sin 5x - 2 \sin 3x + \sin x}{\cos 5x - \cos x} = \tan x$

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

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

$\begin{aligned} \text{LHS} & = \dfrac{\sin 5x - 2 \sin 3x + \sin x}{\cos 5x - \cos x} \\ & \\ & = \dfrac{2 \sin \left(\dfrac{5x+x}{2}\right) \cos \left(\dfrac{5x-x}{2}\right)-2\sin 3x}{-2\sin \left(\dfrac{5x+x}{2}\right) \sin \left(\dfrac{5x-x}{2}\right)} \\ & \\ & = \dfrac{2 \sin 3x \cos 2x - 2 \sin 3x}{-2 \sin 3x \cdot \sin 2x} \\ & \\ & = \dfrac{2 \sin 3x \left(\cos 2x - 1\right)}{-2\sin 3x \cdot \sin 2x} \\ & \\ & = \dfrac{\cos 2x - 1}{-\sin 2x} \\ & \\ & = \dfrac{-2 \sin^2 x}{-2\sin x \cos x} \qquad \left[\text{Note: }\cos 2\theta - 1 = -2 \sin^2 \theta \; ; \quad \sin 2x = 2 \sin x \cos x\right] \\ & \\ & = \dfrac{\sin x}{\cos x} \\ & \\ & = \tan x = \text{RHS} \end{aligned}$

Trigonometric Functions

Show that $\dfrac{1}{2}\left(\cos 4x + \cos 2x\right) = \cos 3x \cos x$

$\cos 4x = \cos \left(3x + x\right)$ and $\cos 2x = \left(3x - x\right)$ $\cos \left(3x + x\right) = \cos 3x \cos x - \sin 3x \sin x$ $\cos \left(3x - x\right) = \cos 3x \cos x + \sin 3x \sin x$ $ \begin{aligned} \therefore \dfrac{1}{2}\left(\cos 4x + \cos 2x\right) & = \dfrac{1}{2} \left[\cos \left(3x + x\right) + \cos \left(3x - x\right)\right] \\ & = \dfrac{1}{2} \left(\cos 3x \cos x - \sin 3x \sin x + \cos 3x \cos x + \sin 3x \sin x\right) \\ & = \dfrac{1}{2} \times 2 \cos 3x \cos x \\ & = \cos 3x \cos x = \text{RHS} \end{aligned} $

Trigonometric Functions

Prove that $\left(1+\cot \theta - \text{cosec} \theta\right) \left(1+\tan \theta + \sec \theta\right) = 2$

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

Trigonometric Functions

Prove that $\sqrt{\sec^2 \theta + \text{cosec}^2 \theta} = \tan \theta + \cot \theta$

$ \begin{aligned} \text{LHS} = \sqrt{\sec^2 \theta + \text{cosec}^2 \theta} & = \sqrt{\left(1+\tan^2 \theta\right) + \left(1 + \cot^2 \theta\right)} \\ & \left[\text{Note: }1+\tan^2 \theta = \sec^2 \theta; 1+\cot^2 \theta = \text{cosec}^2 \theta\right] \\ & = \sqrt{\tan^2 \theta + 2 + \cot^2 \theta} \\ & = \sqrt{\tan^2 \theta + 2 \tan \theta \cot \theta + \cot^2 \theta} \\ & \left[\text{Note: } \tan \theta \times \cot \theta = 1 \right] \\ & = \sqrt{\left(\tan \theta + \cot \theta\right)^2} \\ & = \tan \theta + \cot \theta = \text{RHS} \end{aligned} $

Trigonometric Functions

Show that $\left(\text{cosec}\theta - \sin \theta\right) \left(\sec\theta - \cos\theta\right) \left(\tan\theta + \cot\theta\right) = 1$

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