Prove that

(i) $\sin 75^{\circ}=\sin \left(90^{\circ}-15^{\circ}\right)$ ............(using $\sin (A-B)=\sin A \cos B-\cos A \sin B)$

$=\sin 90^{\circ} \cos 15^{\circ}-\cos 90^{\circ} \sin 15^{\circ}$

$=1 \cdot \cos 15^{\circ}-0 \cdot \sin 15^{\circ}$

$=\cos 15^{\circ}$

$\operatorname{Cos} 15^{\circ}=\cos \left(45^{\circ}-30^{\circ}\right)$ .............. (using $\cos (A-B)=\cos A \cos B+\sin A \sin B$ )

$=\cos 45^{\circ} \cdot \cos 30^{\circ}+\sin 45^{\circ} \cdot \sin 30^{\circ}$

$=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}$

$=\frac{\sqrt{3}+1}{2 \sqrt{2}} \cdot 1 \Rightarrow \frac{\sqrt{3}+1}{2 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin 75^{\circ}=\cos 15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$

(ii) $\frac{\cos 135^{\circ}-\cos 120^{\circ}}{\cos 135^{\circ}+\cos 120^{\circ}}=\frac{\cos \left(180^{\circ}-45^{\circ}\right)-\cos \left(180^{\circ}-60^{\circ}\right)}{\cos \left(180^{\circ}-45^{\circ}\right)+\cos \left(180^{\circ}-60^{\circ}\right)}$ (using $\sin \left(180^{\circ}-x\right)$

$=\sin x)$

(using $\cos \left(180^{\circ}-x\right)=-\cos x$ )

$=\frac{-\cos 45^{\circ}-\left(-\cos 60^{\circ}\right)}{-\cos 45^{\circ}+\left(-\cos 60^{\circ}\right)}$

$=\frac{\cos 60^{\circ}-\cos 45^{\circ}}{-\left(\cos 60^{\circ}+\cos 45^{\circ}\right)}$

$=-\frac{\frac{1}{2}-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}+\frac{1}{2}} \Rightarrow-\frac{\frac{1-\sqrt{2}}{2}}{\frac{\sqrt{2}+1}{2}}=-\frac{1-\sqrt{2}}{\sqrt{2}+1} \cdot \frac{(-\sqrt{2}+1)}{(-\sqrt{2}+1)}$

$=-\frac{-\sqrt{2}+1+2-\sqrt{2}}{-2+\sqrt{2}-\sqrt{2}+1} \Rightarrow-\frac{-2 \sqrt{2}+3}{-1}=3-2 \sqrt{2}$

(iii) $\tan 15^{\circ}+\cot 15^{\circ}=$

First, we will calculate $\tan 15^{\circ}$,

$\tan 15^{\circ}=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}$ ………………….(1)

$\left[\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}, \sin 15^{\circ}=\sin \left(45^{\circ}-30^{\circ}\right)\right.$

$=\sin 45^{\circ} \cdot \cos 30^{\circ}-\cos 45^{\circ} \cdot \sin 30^{\circ}=\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \cdot \frac{1}{2}$

$=\frac{\sqrt{3}-1}{2 \sqrt{2}}$

$\tan 15^{\circ}=\frac{\frac{\sqrt{3}-1}{2 \sqrt{2}}}{\frac{\sqrt{3}+1}{2 \sqrt{2}}} \Rightarrow \frac{\sqrt{3}-1}{\sqrt{3}+1}$ and $\cot 15^{\circ}$

$=\frac{1}{\tan 15^{\circ}} \frac{1}{\frac{\sqrt{3}-1}{\sqrt{3}+1}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Putting in eq(1),

$\tan 15^{\circ}+\cot 15^{\circ}=\frac{\sqrt{3}-1}{\sqrt{3}+1}+\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$=\frac{(\sqrt{3}-1)^{2}+(\sqrt{3}+1)^{2}}{3-1} \frac{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}}{2}$

$=\frac{8}{2}=4$