Prove the following trigonometric identities.

(i) We have,

$\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}=\frac{\sqrt{\sec \theta-1}}{\sqrt{\sec \theta+1}}+\frac{\sqrt{\sec \theta+1}}{\sqrt{\sec \theta-1}}$

$=\frac{\sqrt{\sec \theta-1} \sqrt{\sec \theta-1}+\sqrt{\sec \theta+1} \sqrt{\sec \theta+1}}{\sqrt{\sec \theta+1} \sqrt{\sec \theta-1}}$

$=\frac{(\sqrt{\sec \theta-1})^{2}+(\sqrt{\sec \theta+1})^{2}}{\sqrt{(\sec \theta-1)(\sec \theta+1)}}$

$=\frac{\sec \theta-1+\sec \theta+1}{\sqrt{\sec ^{2} \theta-1}}$

$=\frac{2 \sec \theta}{\sqrt{\tan ^{2} \theta}}$

$=\frac{2 \sec \theta}{\tan \theta}$

$=\frac{2 \frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}$

$=2 \frac{1}{\sin \theta}$

$=2 \operatorname{cosec} \theta$

(ii) We have,

$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{\sqrt{1+\sin \theta}}{\sqrt{1-\sin \theta}}+\frac{\sqrt{1-\sin \theta}}{\sqrt{1+\sin \theta}}$

$=\frac{\sqrt{1+\sin \theta} \sqrt{1+\sin \theta}+\sqrt{1-\sin \theta} \sqrt{1-\sin \theta}}{\sqrt{1-\sin \theta} \sqrt{1+\sin \theta}}$

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

$=\frac{1+\sin \theta+1-\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$

$=\frac{2}{\sqrt{\cos ^{2} \theta}}$

$=\frac{2}{\cos \theta}$

$=2 \sec \theta$

(iii) We have,

$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}+\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\frac{\sqrt{1+\cos \theta}}{\sqrt{1-\cos \theta}}+\frac{\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}}$

$=\frac{\sqrt{1+\cos \theta} \sqrt{1+\cos \theta}+\sqrt{1-\cos \theta} \sqrt{1-\cos \theta}}{\sqrt{1-\cos \theta} \sqrt{1+\cos \theta}}$

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

$=\frac{1+\cos \theta+1-\cos \theta}{\sqrt{1-\cos ^{2} \theta}}$

$=\frac{2}{\sqrt{\sin ^{2} \theta}}$

$=\frac{2}{\sin \theta}$

$=2 \operatorname{cosec} \theta$

(iv) We have,

$\frac{\sec \theta-1}{\sec \theta+1}=\frac{\frac{1}{\cos \theta}-1}{\frac{1}{\cos \theta}+1}$

$=\frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{1+\cos \theta}{\cos \theta}}$

$=\frac{1-\cos \theta}{1+\cos \theta}$

Multiplying both the numerator and the denominator by $(1+\cos \theta)$, we have

$\frac{\sec \theta-1}{\sec \theta+1}=\frac{(1-\cos \theta)(1+\cos \theta)}{(1+\cos \theta)(1+\cos \theta)}$

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

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

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

(v) We have,

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

Multiplying both the numerator and the denominator by $\sin \theta+(1-\cos \theta)$, we have

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

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

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

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

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

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

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

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

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

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

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