Question

Solve the problem!! ​

Answer 1

$\huge \bf{ \mid{ \overline{ \underline \red{Answer - }} \mid}}$

We have

LHS =$\frac{1}{ (\sec\theta - \tan \theta) } - \frac{1}{ \cos\theta } \\ = \frac{1}{ (\sec \theta - \tan\theta) } \times \frac{( \sec \theta + \tan \theta) }{ (\sec \theta + \tan \theta) } - \sec\theta \\ = \frac{( \sec\theta + \tan \theta) }{( { \sec }^{2} \theta - { \tan }^{2} \theta) } - \sec\theta \\ = ( \sec\theta + \tan \theta) - \sec \theta \: ( \because \: { \sec }^{2} \theta - { \tan}^{2} \theta = 1) \\ = \tan \theta$

RHS =$\frac{1}{ \cos\theta } - \frac{1}{ (\sec\theta + \tan \theta) } \\ = \sec \theta - \frac{1}{( \sec \theta + \tan \theta) } \times \frac{( \sec \theta - \tan \theta) }{( \sec \theta - \tan \theta) } \\ = \sec \theta - \frac{( \sec \theta - \tan \theta) }{ { (\sec}^{2} \theta - { \tan }^{2} \theta) } \\ = \sec \theta - ( \sec \theta - \tan \theta) \: \: \: ( \because \: { \sec}^{2} \theta - { \tan}^{2} \theta = 1) \\ = \tan \theta$

$\therefore$LHS =RHS

$\boxed{ \boxed{ \bold{proved}}}$

Answer 2

$\large{Refer\:To\:Attachment}$

$\huge{Divya}$❤