Vivekanand Vellanki
0

It helps to know what you tried so that we can help you appropriately.

I tried expanding the left hand side and simplifying and was able to solve it.

\frac{\left(1+\sin\theta-\cos\theta\right)^2}{\left(1+\sin\theta+\cos\theta\right)^2}=\frac{\left(1+\sin\theta-\cos\theta+\sin\theta+\sin^2\theta-\sin\theta\cos\theta-\cos\theta-\sin\theta\cos\theta+\cos^2\theta\right)}{\left(1+\sin\theta+\cos\theta+\sin\theta+\sin^2\theta+\sin\theta\cos\theta+\cos\theta+\sin\theta\cos\theta+\cos^2\theta\right)}

LHS = \frac{1+2\sin\theta-2\cos\theta-2\sin\theta\cos\theta+\sin^2\theta+\cos^2\theta}{1+2\sin\theta+2\cos\theta+2\sin\theta\cos\theta+\sin^2\theta+\cos^2\theta}

LHS = \frac{\left(2+2\sin\theta-2\cos\theta-2\sin\theta\cos\theta\right)}{\left(2+2\sin\theta+2\cos\theta+2\sin\theta\cos\theta\right)} Using \sin^2\theta+\cos^2\theta=\ 1

LHS = \frac{2\left(1+\sin\theta\right)\ -2\cos\theta\left(1+\sin\theta\right)}{2\left(1+\sin\theta\right)\ +2\cos\theta\left(1+\sin\theta\right)}=\frac{\left(1+\sin\theta\right)\left(2-2\cos\theta\right)}{\left(1+\sin\theta\right)\left(2+2\cos\theta\right)}

Dividing the numerator and denominator by \left(1+\sin\theta\right)\cdot2 gives

LHS = \frac{1-\cos\theta}{1+\cos\theta} = RHS