A geometry problem by arghya starx

$\begin{aligned}(\sec A + \tan A)(1-\sin A)&= \left(\dfrac{1+\sin A}{\cos A}\right)(1-\sin A)\\ &= \dfrac{1-\sin^2 A}{\cos A}\\ &= \dfrac{\cos^2 A}{\cos A}\\ &= \boxed{\cos A}\end{aligned}$

$(sec(A)+tan(A))(1-sin(A))\\ (\frac { 1 }{ cos(A) } +\frac { sin(A) }{ cos(A) } )(1-sin(A)\\ \frac { 1 }{ cos(A) } -\frac { sin(A) }{ cos(A) } +\frac { sin(A) }{ cos(A) } -\frac { { sin }^{ 2 }(A) }{ cos(A) } \\ \frac { 1-{ sin }^{ 2 }(A) }{ cos(A) } \\ \frac { { cos }^{ 2 }(A) }{ cos(A) } \\ cos(A)$

(secA+tanA)(1-sinA)=(1-sinA)/(secA-tanA)=(1-sinA)/(secA(1-sinA))=1/secA=cosA

SIMPLY PUT THE VALUES AND FIND ANSWER IN OPTIONS SAY, A=O Sec O=1, Tan 0=0, Sin 0 =O THEN, (Sec A + tan A)(1- Sin A) = ( 1 +0)(1-o) =(1)(1) =1 and we all know that cos 0 = 1 so the answer is cosA

(1/cosA +sinA/cosA)(1-sinA) =(1+sinA)(1-sinA)/cosA =(1-sin^2A)/cosA =cos^2A/cosA= cosA

(secA+tanA)(1-sinA)

Multiply this expression by (1+sinA)/(1+sinA)

=[(secA+tanA)(cos^2A)]/(1+sinA)

=(cosA+sinAcosA)/(1+sinA)

=[cosA(1+sinA)]/(1+sinA)

=cosA

=(1/cosa +sinA/cosA)(1-sinA) Taking LCM,we get =(1+sinA/cosA)(1-sinA)

=(1-sin^2A)/cosa

=Cos^2A/cosa

=CosA

So simple,ask sumthing cumples

(sec A + tan A)(1- sin A)= sec A + tan A - tan A - tan A sin A sec A - tan A sin A sec A - sin A . (sin A /cos A) sec A (1- sin ^2 A) sec A . cos ^2 A = cos A