Prove the following trigonometric identities.

We need to prove $\frac{1-\cos A}{1+\cos A}=(\cot A-\operatorname{cosec} A)^{2}$

Now, rationalising the L.H.S, we get

$\frac{1-\cos A}{1+\cos A}=\left(\frac{1-\cos A}{1+\cos A}\right)\left(\frac{1-\cos A}{1-\cos A}\right)$

$=\frac{(1-\cos A)^{2}}{1-\cos ^{2} A}$ $\left(\right.$ Using $\left.a^{2}-b^{2}=(a+b)(a-b)\right)$

$=\frac{1+\cos ^{2} A-2 \cos A}{\sin ^{2} A}$ (Using $\sin ^{2} \theta=1-\cos ^{2} \theta$ )

$=\frac{1}{\sin ^{2} A}+\frac{\cos ^{2} A}{\sin ^{2} A}-\frac{2 \cos A}{\sin ^{2} A}$

Using $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$ and $\cot \theta=\frac{\cos \theta}{\sin \theta}$, we get

$\frac{1}{\sin ^{2} A}+\frac{\cos ^{2} A}{\sin ^{2} A}-\frac{2 \cos A}{\sin ^{2} A}=\operatorname{cosec}^{2} A+\cot ^{2} A-2 \cot A \operatorname{cosec} A$

$=(\cot A-\operatorname{cosec} A)^{2}$ $\left(\right.$ Using $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right)$

Hence proved.