Question

(secA-tanA)²(1+sinA)=1-sinA

Answer 1

this is the correct answer.
I hope you understand

Answer 2

Step-by-step explanation:

Given question is to prove

$(secA-tanA)^2(1+sinA)=1-sinA$

$LHS\ =\ (secA-tanA)^2(1+sinA)$

        $=\ (sec^2A+tan^2A-2secA.tanA)(1+sinA)$

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

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

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

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

        $=\ 1-sinA$

$RHS\ =\ 1-sinA$

Hence, we can see that

LHS = RHS

Hence the given equations are proved.