Prove the following trigonometric identities.

We need to prove $(\sec A-\tan A)^{2}=\frac{1-\sin A}{1+\sin A}$

Here, we will first solve the L.H.S.

Now, using $\sec \theta=\frac{1}{\cos \theta}$ and $\tan \theta=\frac{\sin \theta}{\cos \theta}$, we get

$(\sec A-\tan A)^{2}=\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)^{2}$

$=\left(\frac{1-\sin A}{\cos A}\right)^{2}$

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

Further using the property $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get

So,

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

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

$=\frac{1-\sin A}{1+\sin A}$

Hence proved.