# Prove that:

Question:

Prove that:

(i) $\frac{\sin A+\sin 3 A}{\cos A-\cos 3 A}=\cot A$

(ii) $\frac{\sin 9 A-\sin 7 A}{\cos 7 A-\cos 9 A}=\cot 8 \mathrm{~A}$

(iii) $\frac{\sin A-\sin B}{\cos A+\cos B}=\tan \frac{A-B}{2}$

(iv) $\frac{\sin A+\sin B}{\sin A-\sin B}=\tan \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)$

(v) $\frac{\cos A+\cos B}{\cos B-\cos A}=\cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)$

Solution:

(i) Consider LHS :

$\frac{\sin A+\sin 3 A}{\cos A-\cos 3 A}$

$=\frac{2 \sin \left(\frac{A+3 A}{2}\right) \cos \left(\frac{A-3 A}{2}\right)}{2 \sin \left(\frac{A+3 A}{2}\right) \sin \left(\frac{3 A-A}{2}\right)}$

$\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right.$, and $\left.\cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) c \cos \left(\frac{B-A}{2}\right)\right\}$

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

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

$=\cot A$

$=$ RHS

Hence, LHS $=$ RHS.

(ii) Consider LHS :

$\frac{\sin 9 A-\sin 7 A}{\cos 7 A-\cos 9 A}$

$=\frac{2 \sin \left(\frac{9 A-7 A}{2}\right) \cos \left(\frac{9 A+7 A}{2}\right)}{2 \sin \left(\frac{7 A+9 A}{2}\right) \sin \left(\frac{9 A-7 A}{2}\right)}$

$\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right.$ and $\left.\cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{B-A}{2}\right)\right]$

$=\frac{\sin A \cos 8 A}{\sin 8 A \sin A}$

$=\cot 8 A$

$=$ RHS

Hence, LHS $=$ RHS.

(iii) Consider LHS :

$\frac{\sin A-\sin B}{\cos A+\cos B}$

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

$\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right.$ and $\left.\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$

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

$=\tan \left(\frac{A-B}{2}\right)$

$=$ RHS

Hence, LHS $=$ RHS.

(iv) Consider LHS :

$\frac{\sin A+\sin B}{\sin A-\sin B}$

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

$\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right.$, and $\left.\sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right\}$

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

$=\tan \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)$

$=$ RHS

Hence, LHS $=$ RHS.

(v) Consider LHS :

$\frac{\cos A+\cos B}{\cos B-\cos A}$

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

$\left[\because \cos A+\cos B=2 \cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right.$ and $\left.\cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{B-A}{2}\right)\right]$

$=\frac{\cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}{\sin \left(\frac{A+B}{2}\right) s \operatorname{in}\left(\frac{A-B}{2}\right)}$

$=\cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)$

$=$ RHS

Hence, LHS $=$ RHS.