Prove the following trigonometric identities.

Solution:

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

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

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

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

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

So,

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

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

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

$=\sec A+\tan A$

Hence proved.

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

Here, rationaliaing the L.H.S, we get

$\sqrt{\frac{1-\cos A}{1+\cos A}}=\sqrt{\frac{1-\cos A}{1+\cos A}} \times \sqrt{\frac{1-\cos A}{1-\cos A}}$

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

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

So,

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

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

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

$=\operatorname{cosec} A-\cot A$

Hence proved.