Question

prove L.H.S = R.H.S.

Answer 1

we have to prove that :

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

On taking LHS :

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

On taking RHS :

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

Answer 2

Use the following identity:

$\tan(x + y) = \frac{ \tan(x) + \tan(y) }{1 - \tan(x) \tan(y) }$
Plug in x = pi/4 and y = A/2.
$\tan( \frac{\pi}{4} + \frac{a}{2} ) = \frac{ \tan( \frac{\pi}{4} ) + \tan( \frac{a}{2} ) }{1 - \tan( \frac{\pi}{4} ) \times \tan( \frac{a}{2} ) } \\ = \frac{1 + \tan( \frac{a}{2} ) }{1 - \tan( \frac{a}{2} ) } \\ = \frac{1 + \tan( \frac{a}{2} ) }{1 - \tan( \frac{a}{2} ) } \times \frac{1 + \tan( \frac{a}{2} ) }{1 + \tan( \frac{a}{2} ) } \\ = \frac{ {(1 + \tan( \frac{a}{2}) ) }^{2} }{1 - \tan ^{2} (\frac{a}{2} ) }$
Now,

observe that,

$\cos(2x) = \frac{1 - \tan^{2} (x) }{1 + \tan^{2} (x) } \\ \cos(a) = \frac{1 - \tan^{2} ( \frac{a}{2} ) }{1 + \tan ^{2} ( \frac{a}{2} ) } \\ \cos(a) \times (1 + \tan ^{2} ( \frac{a}{2} ) ) = 1 - \tan^{2} ( \frac{a}{2} )$
Thus the expression becomes,

$\frac{(1 + \tan^{2}( \frac{a}{2} )) ^{2} }{ \cos(a) \times (1 + \tan^{2} ( \frac{a}{2} )) } \\ = \frac{1 + \tan ^{2} ( \frac{a}{2} ) + 2 \tan( \frac{a}{2} ) }{ \cos(a) \times (1 + \tan ^{2} ( \frac{a}{2} )) } \\ = \frac{1 + \tan ^{2} ( \frac{a}{2} )}{\cos(a) \times (1 + \tan ^{2} ( \frac{a}{2} ))} + \frac{2 \tan( \frac{a}{2} ) }{\cos(a) \times (1 + \tan ^{2} ( \frac{a}{2} ))} \\ = \frac{1 }{ \cos(a) } + \frac{1}{ \cos(a) } \times \frac{2 \tan( \frac{a}{2} ) }{1 + \tan ^{2} ( \frac{a}{2} ) } \\ = \frac{1}{ \cos(a) } + \frac{1}{ \cos(a) } \times 2 \sin( \frac{a}{2} ) \cos( \frac{a}{2} ) \\ = \frac{1}{ \cos(a) } + \frac{ \sin(a) }{ \cos(a) } \\ = \frac{1 + \sin(a) }{ \cos(a) } \\ = \frac{(1 + \sin(a))(1 - \sin(a)) }{( \cos(a) (1 - \sin(a) ) } \\ = \frac{ \cos^{2} (a) }{ \cos(a)(1 - \sin(a)) } \\ = \frac{ \cos(a) }{1 - \sin(a) }$


The other identities I have used are:

$\sin(2x) = 2 \sin(x) \cos(x) \\ \cos(2x) = \cos^{2} (x) - \sin ^{2} (x)$