Question

Prove that - TAN theta - cos theta / sin theta×cos theta = tan^2 theta - cos^ theta

Answer 1

Given      $\dfrac{\tan \theta - \cot \theta}{\sin \theta \cdot \cos\theta} = \tan^2 \theta - \cot^2 \theta$

To Find : Prove

Solution:

$\dfrac{\tan \theta - \cot \theta}{\sin \theta \cdot \cos\theta} = \tan^2 \theta - \cot ^2 \theta$

RHS = $\tan^2 \theta - \cot ^2 \theta$

$= \dfrac{\sin^2 \theta}{\cos^2 \theta} - \dfrac{\cos^2 \theta}{\sin^2 \theta}$

$= \dfrac{\sin^4 \theta - \cos^4 \theta}{\cos^2 \theta\sin^2 \theta}$

$= \dfrac{(\sin^2 \theta + \cos^2 \theta)(\sin^2 \theta - \cos^2 \theta)}{\cos^2 \theta\sin^2 \theta}$

$= \dfrac{1(\sin^2 \theta - \cos^2 \theta)}{\cos^2 \theta \sin^2 \theta}$

$= \dfrac{ \sin^2 \theta - \cos^2 \theta }{\cos \theta \sin \theta\cos \theta \sin \theta}$

$=\dfrac{1}{\cos \theta \sin \theta} \left( \dfrac{ \sin^2 \theta - \cos^2 \theta }{\\cos \theta \sin \theta} \right)$

$=\dfrac{1}{\cos \theta \sin \theta} \left( \dfrac{ \sin^2 \theta }{\\cos \theta \sin \theta} - \dfrac{ \cos^2 \theta }{\\cos \theta \sin \theta} \right)$

$=\dfrac{1}{\cos \theta \sin \theta} \left( \dfrac{\sin \theta }{\cos \theta } - \dfrac{ \cos \theta }{\sin \theta} \right)$

$=\dfrac{1}{\cos \theta \sin \theta} \left(\tan \theta } -\cot \theta} \right)$

$=\dfrac{\tan \theta -\cot \theta}{\cos \theta \sin \theta}$

= LHS

 $\dfrac{\tan \theta - \cot \theta}{\sin \theta \cdot \cos\theta} = \tan^2 \theta - \cot^2 \theta$

QED

HENCE PROVED

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