Question

Prove that

$\ \Big(secA-tanA\Big)^2= \dfrac{1-sinA}{1+sinA}$
​

Answer 1

To Prove :-

• $\sf \ \Big(secA-tanA\Big)^2= \dfrac{1-sinA}{1+sinA}$

Proof :-

$\sf\ \ \ \ L.H.S$

$:\implies \sf \ \Big(secA-tanA\Big)^2$

$:\implies \sf \bigg(\dfrac{1}{cosA}-\dfrac{sinA}{cosA}\bigg)^2$

$\sf\bigg[\therefore\ secA=\dfrac{1}{cosA}$$\sf\ tanA= \dfrac{sinA}{cosA}\bigg]$

$:\implies \sf \bigg(\dfrac{1-sinA}{cosA}\bigg)^2$

$:\implies \sf \dfrac{\Big(1-sinA\Big)^2}{cos^2A}$

$\sf\Big[(1-sinA)^2=(1-sinA)(1-sinA)\Big]$

$:\implies \sf \dfrac{\Big(1-sinA\Big)\Big(1-sinA\Big)}{1-sin^2A}$

$\sf\Big[\therefore\ cos^2A=1-sin^2A \Big]$

$:\implies \sf \dfrac{\Big(1-sinA\Big) \Big(1-sinA\Big)}{\Big(1+sinA\Big)\Big(1-sinA\Big)}$

$\sf\Big[\therefore\ a^2-b^2=(a+b)(a-b)\Big]$

$:\implies\sf \dfrac{1-sinA}{1+sinA}$

$\sf\ \ \ \ R.H.S$

Answer 2

Answer:

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Step-by-step explanation:

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