Question

$\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} = \dfrac{1-cos}{1+cos}$ ​

Answer 1

$\huge\underbrace\mathrm\purple{Answer}$

I=integ.of 1/(1-cosa.cosx). dx.

I=integ.of 1/[1-cosa(1-tan^2x/2)/(1+tan^2x/2)].dx

I=integ.of (1+tan^2x/2)/[1+tan^2x/2-cosa+cosa.tan^2x/2].dx

I=integ.of sec^2x/2.dx/[(1+cosa).tan^2x/2+(1-cosa)]

Let tan x/2=p

1/2.sec^2x/2.dx=dp. or. sec^2x/2.dx=2.dp

I=integ.of 2.dp/[(1-cosa)+(1+cosa).p^2]

I=integ.of 2.dp/[2.sin^a/2+2.cos^2a/2..p^2]

I=integ.of. dp/[sin^2a/2+ cos^2a/2.p^2]

I=integ.of. cosec^2a/2.dp/[1 + ( p.cot a/2)^2]

I=cosec^a/2.(1/cot a/2).tan^-1(p.cot a/2). + C

I=(1/sin^2a/2).(sin a/2/cos a/2).tan^-1(cot a/2.tan x/2). + C.

I=2.cosec a. tan^-1(cot a/2.tan x/2). + C. Answer.