Question

Prove that:

(i) $\frac{1}{\sin \left(x-a\right)\sin \left(x-b\right)}=\frac{\cot \left(x-a\right)-\cot \left(x-b\right)}{\sin \left(a-b\right)}$
(ii) $\frac{1}{\sin \left(x-a\right)\cos \left(x-b\right)}=\frac{\cot \left(x-a\right)+\tan \left(x-b\right)}{\cos \left(a-b\right)}$
(iii) $\frac{1}{\cos \left(x-a\right)\cos \left(a-b\right)}=\frac{\tan \left(x-b\right)-\tan \left(x-a\right)}{\sin \left(a-b\right)}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{RHS}\hspace{0.17em}=\frac{\cot \left(x-a\right)-\cot (x-b)}{\sin (a-b)} \\ =\frac{{\displaystyle \frac{\cos (x-a)}{\sin (x-a)}}-{\displaystyle \frac{\cos (x-b)}{\sin (x-b)}}}{\sin (a-b)} \\ =\frac{\sin (x-b)\cos (x-a)-\sin (x-a)\cos (x-b)}{\sin (x-a)\sin (x-b)\sin (a-b)} \\ =\frac{\sin (x-b-x+a)}{\sin (x-a)\sin (x-b)\sin (a-b)} \\ =\frac{\sin (a-b)}{\sin (x-a)\sin (x-b)\sin (a-b)} \\ =\frac{1}{\sin (x-a)\sin (x-b)} \\ =\mathrm{LHS} \\ \mathrm{Hence}\text{p}\mathrm{roved}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{RHS}=\frac{\cot (x-a)+\tan (x-b)}{\cos (a-b)} \\ =\frac{{\displaystyle \frac{\cos (x-a)}{\sin (x-a)}}+{\displaystyle \frac{\sin (x-b)}{\cos (x-b)}}}{\cos (a-b)} \\ =\frac{\cos (x-b)\cos (x-a)+\sin (x-a)\sin (x-b)}{\cos (a-b)\sin (x-a)\cos (x-b)} \\ =\frac{\cos (x-b-x+a)}{\cos (a-b)\sin (x-a)\cos (x-b)}(\mathrm{Using}\cos (A-B)=\cos A\cos bB+\sin A\sin B) \\ =\frac{\cos (a-b)}{\cos (a-b)\sin (x-a)\cos (x-b)} \\ =\frac{1}{\sin (x-a)\cos (x-b)} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{RHS}=\frac{\tan (x-b)-\tan (x-a)}{\sin (a-b)} \\ =\frac{{\displaystyle \frac{\sin (x-b)}{\cos (x-b)}}-{\displaystyle \frac{\sin (x-a)}{\cos (x-a)}}}{\sin (a-b)} \\ =\frac{\sin (x-b)\cos (x-a)-\sin (x-a)\cos (x-b)}{\sin (a-b)\cos (x-a)\cos (x-b)} \\ =\frac{\sin (x-b-x+a)}{\sin (a-b)\cos (x-a)\cos (x-b)}(\mathrm{Using}\sin (A-B)=\sin A\cos B-\cos A\sin B) \\ =\frac{\sin (a-b)}{\sin (a-b)\cos (x-a)\cos (x-b)} \\ =\frac{1}{\cos (x-a)\cos (x-b)} \\ =\mathrm{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$