Question

1/1+sinA + 1/1-sinA = 2secA

Answer 1

/* There is a mistake in RHS part */

$Given \: \frac{1}{(1+sin A )} + \frac{1}{(1-sin A)}$

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

$= \frac{ 2}{ 1^{2} - sin^{2} A }$

$= \frac{2}{ cos^{2} A }$

$= 2 sec^{2} A$

$= RHS$

Therefore.,

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

$\green {= 2 sec^{2} A }$

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Answer 2

To Prove :

$\sf \dfrac{1}{1+sinA}+\dfrac{1}{1-sinA}=2.sec^2A$

Identities Applied :

$\bullet\ \; \sf (a+b)(a-b)=a^2-b^2\\\\\bullet\ \; \sf cos^2\theta=1-sin^2\theta\\\\\bullet\ \; sec\theta=\dfrac{1}{cos\theta}$

Proof :

Take LHS

$\to \sf \dfrac{1}{1+sinA}+\dfrac{1}{1-sinA}\\\\\to \sf \dfrac{(1-sinA)+(1+sinA)}{(1+sinA)(1-sinA)}\\\\\to \sf \dfrac{2}{1-sin^2A}\\\\\to \sf \dfrac{2}{cos^2A}\\\\\to \sf 2.sec^2A\\\\\to \sf RHS$

Hence Proved