Pls solve this question

For simplicity taking θ=x

To Prove {(cotx - cosecx)}^{2} = \frac{1 - cosx}{1 + cosx} Consider {(cotx - cosecx)}^{2} = {(\frac{cosx}{sinx} - \frac{1}{sinx})}^{2} = {(\frac{cosx - 1}{sinx})}^{2} = \frac{{(cosx - 1)}^{2}}{1 - \cos^{2} x} = \frac{{(cosx - 1)}^{2}}{- (\cos^{2} x - 1)} = \frac{(cosx - 1) (cosx - 1)}{- (cosx - 1) (cosx + 1)} = \frac{1 - cosx}{1 + cosx} = RHS Note : \cos^{2} x + \sin^{2} x = 1