Question

(se A-tana)^2=1-sinA/1+sinA

Answer 1

Given :-

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

• We have to prove the given trigonometric identity

Now, taking LHS,

$\dashrightarrow\sf \bigg(\dfrac{1}{cosA}-\dfrac{sinA}{cosA}\bigg)^2 \ \ \ \ \ \bigg[\therefore\ secA=\dfrac{1}{cosA}\ \ ; \ \ tanA= \dfrac{sinA}{cosA}\bigg]\\ \\ \\ \dashrightarrow\sf \bigg(\dfrac{1-sinA}{cosA}\bigg)^2\\ \\ \\ \dashrightarrow\sf \dfrac{\Big(1-sinA\Big)^2}{cos^2A}\ \ \ \ \ \ \Big[(1-sinA)^2=(1-sinA)(1-sinA)\Big]\\ \\ \\ \dashrightarrow\sf \dfrac{\Big(1-sinA\Big)\Big(1-sinA\Big)}{1-sin^2A}\ \ \ \ \ \Big[\therefore\ cos^2A=1-sin^2A \Big]\\ \\ \\ \dashrightarrow\sf \dfrac{\Big(1-sinA\Big) \Big(1-sinA\Big)}{\Big(1+sinA\Big)\Big(1-sinA\Big)}\ \ \ \ \ \ \Big[\therefore\ a^2-b^2=(a+b)(a-b)\Big]\\ \\ \\ \dashrightarrow\sf \dfrac{1-sinA}{1-sinA}\ \ will \ be \ cancelled\\ \\ \\ :\implies\sf \dfrac{1-sinA}{1+sinA}$

$\bullet\sf \ R.H.S \longrightarrow \dfrac{1-sinA}{1+sinA}\\ \\ \\ \bullet\sf L.H.S \longrightarrow \dfrac{1-sinA}{1+sinA}\\ \\ \\ \sf\ \ \ \ L.H.S= R.H.S\ \ \ \ (Hence \ Proved !!)$