Question

prove that square root of 1+ sin a/1- sin a+ square root of 1- sin a/1+ sin a=2 sec a​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

$\sqrt{\dfrac{1 + sinA}{1 - sinA}} + \sqrt{\dfrac{1 - sinA}{1 + sinA}} = 2secA$

Solution :-

Taking L.H.S :-

$\sqrt{\dfrac{1 + sinA}{1 - sinA}} + \sqrt{\dfrac{1 - sinA}{1 + sinA}}$

Rationalising the denominator :-

$=\sqrt{\dfrac{1 + sinA}{1 - sinA}\times\dfrac{1 + sinA}{1+sinA}} + \sqrt{\dfrac{1 - sinA}{1 + sinA}\times \dfrac{1-sinA}{1-sinA}}$

$= \sqrt{\dfrac{(1+sinA)^2}{1-sin^2A}} +\sqrt{\dfrac{(1-sinA)^2}{1-sin^2A}}$

$= \sqrt{\dfrac{(1+sinA)^2}{cos^2A}} + \sqrt{\dfrac{(1-sinA)^2}{cos^2A}}$

$Since , sin^2A +cos^2A = 1 \implies cos^2A = 1-sin^2A$

$= \sqrt{\bigg(\dfrac{1+sinA}{cosA}\bigg)^2}+ \sqrt{\bigg(\dfrac{1-sinA}{cosA}\bigg)^2}$

Cancelling both the square and the square root we get :-

$= \dfrac{1+sinA}{cosA}+\dfrac{1-sinA}{cosA}$

$= \dfrac{1+sinA+1-sinA}{cosA}$

$= \dfrac{2}{cosA}$

$= 2secA$

$Since, \dfrac{1}{cosA} = secA$

Hence proved.

Know more :-

Trigonometric identities :-

1) sin^2A + cos^2A = 1

2) sec^2A - tan^2A = 1

3) csc^2A - cot^2A = 1