Prove the following trigonometric identities.

(i) We have to prove $\frac{\cos A \operatorname{cosec} A-\sin A \sec A}{\cos A+\sin A}=\operatorname{cosec} A-\sec A$

So,

$\frac{\cos A \operatorname{cosec} A-\sin A \sec A}{\cos A+\sin A}=\frac{\cos A \frac{1}{\sin A}-\sin A \frac{1}{\cos A}}{\cos A+\sin A}$

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

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

$=\frac{(\cos A-\sin A)(\cos A+\sin A)}{\sin A \cos A(\cos A+\sin A)}$

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

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

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

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

Hence proved.

(ii) We have to prove $\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\operatorname{cosec} A+\cot A-1}=1$

We know that, $\sin ^{2} A+\cos ^{2} A=1$

So,

$\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\operatorname{cosec} A+\cot A-1}$

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

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

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

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

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

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

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

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

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

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

$=\frac{2 \sin A \cos A}{2 \sin A \cos A}$

$=1$

Hence proved.