Question

hello every one any one can try this Q

Answer 1

We have to prove that :

$\frac{1 + \sin\alpha }{1 - \sin\alpha } = { \tan}^{2} ( \frac{\pi}{4} + \frac{ \alpha }{2} )$

On taking LHS :

$\frac{1 + \sin\alpha }{1 - \sin\alpha } \\ \\ = > \frac{ { \sin }^{2} \frac{ \alpha }{2} + { \cos }^{2} \frac{ \alpha }{2} + 2 \sin \frac{ \alpha }{2} \cos \frac{ \alpha }{2} }{ { \sin }^{2} \frac{ \alpha }{2} + { \cos }^{2} \frac{ \alpha }{2} - 2 \sin \frac{ \alpha }{2} \cos \frac{ \alpha }{2} } \\ \\ = > \frac{ {( \sin \frac{ \alpha }{2} + \cos \frac{ \alpha }{2} ) }^{2} }{ {( \sin \frac{ \alpha }{2} - \cos \frac{ \alpha }{2} )}^{2} } \\ \\ \\ = > {( \frac{1 + \cot \frac{ \alpha }{2} }{1 - \cot \frac{ \alpha }{2} }) }^{2}$

On taking RHS :

${ \tan }^{2} ( \frac{\pi}{4} + \frac{ \alpha }{2} ) \\ \\ = > {[ \tan( \frac{\pi}{4} + \frac{ \alpha }{2} ) ]}^{2} \\ \\ = > {[ \tan( \frac{\pi}{2} - \frac{\pi}{4} + \frac{ \alpha }{2} )]}^{2} \\ \\ = > {[ \tan( \frac{\pi}{2} - ( \frac{\pi}{4} - \frac{ \alpha }{2} )) ]}^{2} \\ \\ = > {[ \cot( \frac{\pi}{4} - \frac{ \alpha }{2} )] }^{2} \\ \\ = > {(\frac{ \cot \frac{\pi}{4} + \cot \frac{ \alpha }{2} }{1 - \cot \frac{\pi}{4} \cot \frac{ \alpha }{2} } )}^{2} \\ \\ = > {(\frac{1 + \cot \frac{ \alpha }{2} }{1 - \cot \frac{ \alpha }{2} } )}^{2} \\ \\ As \: we \: know \: that :\: \cot \frac{\pi}{4} = 1 \\ \\ = > LHS = RHS \\ \\ HENCE \: PROVED$