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