Question

prove the given ...
immediately​

Answer 1

To Proof :

$\tt\dagger \frac{1}{ \csc \theta - \cot \theta } - \frac{1}{ \sin \theta} = \frac{1}{ \sin \theta} - \frac{1}{ \csc \theta + \cot \theta }$

Proof :

Let's proof LHS and RHS simultaneously,

$\tt\mapsto \frac{1}{ \csc \theta - \cot \theta } - \frac{1}{ \sin \theta} = \frac{1}{ \sin \theta} - \frac{1}{ \csc \theta + \cot \theta }$

$\tt\mapsto \frac{1}{ \csc \theta - \cot \theta } + \frac{1}{ \csc \theta + \cot \theta } = \frac{1}{ \sin \theta} + \frac{1}{ \sin \theta}$

$\tt\mapsto \frac{ \csc \theta + \cot \theta + \csc \theta - \cot \theta}{ (\csc \theta + \cot \theta) (\csc \theta - \cot \theta) } = \frac{2}{ \sin \theta}$

$\tt\mapsto \frac{2 \csc \theta}{ { \csc}^{2} \theta - { \cot}^{2} \theta } = \frac{2}{ \sin \theta}$

$\tt\mapsto \frac{ \csc \theta}{1} = \frac{1}{ \sin \theta}$

$\tt\mapsto \csc \theta = \csc \theta.$

Hence, LHS = RHS.

$\rule{190}2$

$\boxed{\begin{minipage}{7 cm} Required Formula \\ \\ \sin^2\theta + \cos^2\theta=1 \\ \\ 1+\tan^2\theta = \sec^2\theta \\ \\ 1+\cot^2\theta = {csc}^2 \theta \\ \\ (a + b)(a-b) = a^{2} - b^{2} \\ \\ \frac{1}{\sin\theta}= \csc \theta \end{minipage}}$