Question

Solve the Qno. 15 in the attachment

Standard:- 10

Content Quality Solution Required

❎ Don't Spamming ❎

Answer 1

The answer is in the attachment.


Hope it helps!

Answer 2

$\frac{ \frac{ \sin(a) }{ \cos(a) } }{1 - \frac{ \cos(a) }{ \sin(a) } } + \frac{ \frac{ \cos(a) }{ \sin(a) } }{1 - \frac{ \sin(a) }{ \cos(a) } } \\ \\ \\ \\ => \frac{ \sin ^{2} (a) }{ \cos(a) ( \sin(a) - \cos(a)) } + \frac{ \cos ^{2} (a) }{ \sin(a)( \cos(a) - \sin(a)) } \\ \\ \\ \\ => \frac{1}{ \sin(a) - \cos(a) } ( \frac{ \sin ^{2} (a) }{ \cos(a) } - \frac{ \cos ^{2} (a) }{ \sin(a) } ) \\ \\ \\ \\ => \frac{1}{ \sin(a) - \cos(a) } \times \frac{ \sin ^{3} (a) - { \cos ^{3} (a) } }{ \sin(a) \cos(a) } \\ \\ \\ \\ => \frac{( \sin(a) - \cos(a) )( \sin {a}^{2} (a) + \cos ^{2} (a) - sin(a)cos(a) }{( \sin(a) - \cos(a))(sin(a) \cos(a) ) } \\ \\ \\ \\ => \frac{1 + sin(a)cos(a)}{ \sin(a) \cos(a) } \\ \\ \\ \\ => \frac{1}{sin(a)cos(a)} + 1 \\ \\ \\ \\ => sec(a)cosec(a) + 1$




Hence, proved.