Question

Prove the following identity :tan theta-cot theta/sin theta*cos theta = tan 2θ –cot 2θ

Answer 1

Answer:-

To Prove:

$\sf \ \frac{ \tan \theta - \cot \theta }{ \sin \theta. \cos \theta} = { \tan }^{2} \theta - { \cot }^{2} \theta$

tan² theta - Cot² theta can be written as (tan theta + cot theta)(tan theta - cot theta) since (a² - b² ) = (a + b)(a - b).

$\sf \implies \: \frac{ (\tan \theta - \cot \theta)}{ \sin \theta. \cos \theta} = ( \tan \theta \: + \cot \theta)( \tan \theta - \cot \theta) \\ \\\sf \implies \: \frac{1}{ \sin \theta \cos \theta} = \tan \theta + \cot \theta$

Using tan theta = Sin theta/Cos theta and Cot theta = Cos theta/Sin theta in RHS we get,

$\sf \implies \: \frac{1}{ \sin \theta. \cos \theta} = \frac{ \sin \theta}{ \cos \theta} + \frac{ \cos \theta}{ \sin \theta } \\ \\ \sf \implies \: \frac{1}{ \sin \theta. \cos \theta} = \frac{ { \sin }^{2} \theta + { \cos}^{2} \theta }{ \sin \theta. \cos \theta}$

Using Sin² theta + Cos² theta = 1 in RHS we get,

$\sf \implies \: \frac{1}{ \sin \theta. \cos \theta} = \frac{1}{ \sin \theta . \cos \theta } \\ \\ \implies \sf \large \: LHS = RHS$

Hence, Proved.