Question

Prove that:
(i) $\frac{\sin A+\sin B}{\sin A-\sin B}=\tan \left(\frac{A+B}{2}\right)\cot \left(\frac{A-B}{2}\right)$

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

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

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

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

Answer 1

(i)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin A+\sin B}{\sin A-\sin B} \\ =\frac{2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)}{2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)}\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right),\mathrm{and}\sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\right\} \\ =\frac{\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)}{\sin \left({\displaystyle \frac{A-B}{2}}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)} \\ =\tan \left(\frac{A+B}{2}\right)cot\left(\frac{A-B}{2}\right) \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin A+\sin 3A}{\cos A-\cos 3A} \\ =\frac{2\sin \left(\frac{A+3A}{2}\right)\cos \left(\frac{A-3A}{2}\right)}{2\sin \left(\frac{A+3A}{2}\right)\sin \left(\frac{3A-A}{2}\right)}\left\{\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right),\mathrm{and}\cos A-\cos B=2\sin \left(\frac{A+B}{2}\right)c\mathrm{os}\left({\displaystyle \frac{B-A}{2}}\right)\right\} \\ =\frac{\sin 2A\cos \left(-A\right)}{\sin 2A\sin A} \\ =\frac{\sin 2A\cos A}{\sin 2A\sin A} \\ =\text{c}\text{ot}A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(iii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin 9A-\sin 7A}{\cos 7A-\cos 9A} \\ =\frac{2\sin \left(\frac{9A-7A}{2}\right)\cos \left(\frac{9A+7A}{2}\right)}{2\sin \left(\frac{7A+9A}{2}\right)\sin \left(\frac{9A-7A}{2}\right)}\left[\because \sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\mathrm{and}\cos A-\cos B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{B-A}{2}}\right)\right] \\ =\frac{\sin A\cos 8A}{\sin 8A\sin A} \\ =\mathrm{co}t8A \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(iv)

$$\begin{gathered} \mathrm{Consider}\mathrm{LH}S: \\ \frac{\cos A+\mathrm{co}sB}{\cos B-\cos A} \\ =\frac{2\cos \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}{2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}\left[\because \cos A+\cos B=2\cos \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\mathrm{and}\cos A-\cos B=2\sin \left(\frac{A+B}{2}\right)\mathrm{co}s\left({\displaystyle \frac{B-A}{2}}\right)\right] \\ =\frac{\cos \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}{\sin \left(\frac{A+B}{2}\right)s\mathrm{in}\left(\frac{A-B}{2}\right)} \\ =\mathrm{c}\mathrm{ot}\left(\frac{A+B}{2}\right)\cot \left(\frac{A-B}{2}\right) \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$

(v)
$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \frac{\sin A-\sin B}{\cos A+\cos B} \\ =\frac{2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}{2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}\left[\because \sin A-\sin B=2\sin \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\mathrm{and}\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right] \\ =\frac{\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)} \\ =\tan \left(\frac{A-B}{2}\right) \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{LHS}=\mathrm{RHS}. \end{gathered}$$