Question

cosA/1-sinA=secA+tanA​

Answer 1

Answer:

$L. H. S = \frac{cos A}{1 - sin A } \\ = \frac{cos A}{1 - sin A } \times \frac{ 1 + sin A }{1 + sin A } \\ = \frac{cos A( 1 + sin A )}{1 - sin ^{2} A } \\ = \frac{cos A(1 + sin A)}{cos^{2} A} \\ = \frac{1 +sin A}{cos A } \\ = \frac{1}{cos A} + \frac{sinA }{cos A } \\ = secA + tanA = R. H. S\\ hence \: proved$

Answer 2

Answer:

To prove :- cosA/1-sin A=secA+tanA

Proof:-

LHS= cosA/1-sinA

$\frac{ \cos \: a }{1 - \sin \: a }$

Rationalizing the denominator,

$\frac{cos \: a}{1 - sin \: a} \times \frac{1 + \: sin \: a}{1 + \: sin \: a}$

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

LHS=secA+tanA

RHS=secA+tanA

Therefore,

LHS=RHS........hence proved