Question

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

Step-by-step explanation:

We have,

$\frac{ \cot ^{2} (A)( \sec(A) - 1)}{1 + \sin( A) }$

$= \frac{ \cot ^{2} (A)( \sec(A) - 1)( \sec(A) + 1) }{(1 + \sin( A))( \sec(A) + 1)}$

$= \frac{ \cot ^{2} (A)( \sec^{2} (A) - 1)}{(1 + \sin( A))( \sec(A) + 1)}$

$= \frac{ \cot ^{2} (A). \tan^{2} (A) }{(1 + \sin( A))( \sec(A) + 1)}$

$= \frac{1}{(1 + \sin( A))( \sec(A) + 1)}$

$= \frac{(1 - \sin(A)) }{(1 + \sin( A))(1 - \sin(A)) ( \sec(A) + 1)}$

$= \frac{(1 - \sin(A)) }{(1 - \sin ^{2} (A)) ( \sec(A) + 1)}$

$= \frac{(1 - \sin(A)) }{ \cos ^{2} (A)( \sec(A) + 1)}$

$= \frac{ \sec ^{2} ( A )(1 - \sin(A)) }{ ( \sec(A) + 1)}$