Question

solve this question!!!!!!

Answer 1

Step-by-step explanation:

Note: For the better understanding,I am considering θ as A

$Given:(\frac{1+sinA-cosA}{1+sinA+cosA})^2$

$=(\frac{1+sinA-cosA}{1+sinA+cosA})^2 * (\frac{1+sinA-cosA}{1+sinA-cosA})^2$

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

$=(\frac{1+1+2sinA-2sinAcosA-2cosA}{sin^2A+2sinA+sin^2A})^2$

$=(\frac{2+2sinA-2sinAcosA-2cosA}{2sin^2A+sinA})^2$

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

$=(\frac{2[(1-cosA)(1+sinA)}{2sinA(sinA+1)})^2$

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

$=(\frac{1-cosA}{sinA})^2$

$=(\frac{(1-cosA)^2}{sin^2A})$

$=\frac{(1-cosA)^2}{1-cos^2A}$

$=\frac{(1-cosA)^2}{(1-cosA)(1+cosA)}$

$=\boxed{\frac{1-cosA}{1+cosA}}$

Hope it helps!

Answer 2

Step-by-step explanation:

L.H.S

(1+sinФ-cosФ/1+sinФ+cosФ)²

=(1+sinФ-cosФ)²/(1+sinФ+cosФ)²

=(1+sinФ)²-2(1+sinФ)cosФ+cos²Ф/(1+sinФ)²+2(1+sinФ)cosФ+cos²Ф

=1+2sinФ+sin²Ф-(2+2sinФ)cosФ+cos²Ф/1+ 2sinФ+sin²Ф+(2+2sinФ)cosФ+cos²Ф

=1+2sinФ+sin²Ф+cos²Ф-2cosФ-2sinФcos/1+2sinФ+sin²Ф+cos²Ф+2cosФ+2sinФcosФ

=1+2sinФ+1-2cosФ-2sinФcosФ/1+2sinФ+1+2cosФ+2sinФcosФ

=2+2sinФ-2cosФ-2sinФcosФ/2+2sinФ+2cosФ+2sinФcosФ

=2(1+sinФ)-2cosФ(1+sinФ)/2(1+sinФ)+2cosФ(1+sinФ)

=2(1+sinФ)(1-cosФ)/2(1+sinФ)(1+cosФ)

=1-cosФ/1+cosФ