Question

prove that - tan theta-cos theta / sin theta*cos theta =tan^2 theta - cos ^2 theta.

Answer 1

SOLUTION

TO PROVE

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

PROOF

RHS

$\displaystyle \sf{ = { \tan}^{2} \theta - { \cot}^{2} \theta }$

$\displaystyle \sf{ = \frac{{ \sin}^{2} \theta}{{ \cos}^{2} \theta} - \frac{{ \cos}^{2} \theta}{{ \sin}^{2} \theta}}$

$\displaystyle \sf{ = \frac{{ \sin}^{4} \theta - { \cos}^{4} \theta }{{ \sin}^{2} \theta \: { \cos}^{2} \theta} }$

$\displaystyle \sf{ = \frac{({ \sin}^{2} \theta + \: { \cos}^{2} \theta)({ \sin}^{2} \theta \: - { \cos}^{2} \theta)}{{ \sin}^{2} \theta \: { \cos}^{2} \theta} }$

$\displaystyle \sf{ = \frac{({ \sin}^{2} \theta \: - { \cos}^{2} \theta)}{{ \sin}^{2} \theta \: { \cos}^{2} \theta} }$

$\displaystyle \sf{ = \frac{1}{{ \sin}^{} \theta \: { \cos}^{} \theta} \times \frac{({ \sin}^{2} \theta \: - { \cos}^{2} \theta)}{{ \sin}^{} \theta \: { \cos}^{} \theta} }$

$\displaystyle \sf{ = \frac{1}{{ \sin}^{} \theta \: { \cos}^{} \theta} \times \bigg( \frac{{ \sin}^{2} \theta \: }{{ \sin}^{} \theta \: { \cos}^{} \theta} - \frac{{ \cos}^{2} \theta \: }{{ \sin}^{} \theta \: { \cos}^{} \theta}\bigg) }$

$\displaystyle \sf{ = \frac{1}{{ \sin}^{} \theta \: { \cos}^{} \theta} \times \bigg( \frac{{ \sin}^{} \theta \: }{ \: { \cos}^{} \theta} - \frac{{ \cos}^{} \theta \: }{{ \sin}^{} \theta }\bigg) }$

$\displaystyle \sf{ = \frac{1}{{ \sin}^{} \theta \: { \cos}^{} \theta} \times \bigg( \tan \theta - \cot \theta\bigg) }$

$\displaystyle \sf{ = \frac{ \tan \theta - \cot \theta}{ \sin \theta \cos \theta} }$

= LHS

Hence proved

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Answer 2

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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