Question

Show that:

$\frac{\sin A}{\cot A+\text{cosec}A}=2+\frac{\sin A}{\cot A-\text{cosec}A}$

Answer 1

$LHS=\frac{\sin A}{\cot A+\text{cosec}A}$

$=\frac{\sin A}{\frac{\cos A}{\sin A}+\frac{1}{\sin A}}$

$=\frac{\sin A}{\frac{1+\cos A}{\sin A}}$

$=\sin A\times \frac{\sin A}{1+\cos A}$

$=\frac{{\sin }^{2}A}{1+\cos A}$

$=\frac{1-{\cos }^{2}A}{1+\cos A}$

$=\frac{(1-\cos A)(1+\cos A)}{1+\cos A}$

$=1-\cos A$

$\Rightarrow LHS=1-\cos A$---(1)

$RHS=2+\frac{\sin A}{\cot A-\text{cosec}A}$

$=2+\frac{\sin A}{\frac{\cos A}{\sin A}-\frac{1}{\sin A}}$

$=2+\frac{\sin A}{\frac{\cos A-1}{\sin A}}$

$=2+\sin A\times \frac{\sin A}{\cos A-1}$

$=2+\frac{{\sin }^{2}A}{1-\cos A}$

$=2+\frac{1-{\cos }^{2}A}{\cos A-1}$

$=2+\frac{(1-\cos A)(1+\cos A)}{-(1-\cos A)}$

$=2-(1+\cos A)$

$=2-1-\cos A$

$=1-\cos A$

$\Rightarrow RHS=1-\cos A$---(2)

From (1) and (2)

$LHS=RHS$

$\therefore \frac{\sin A}{\cot A+\text{cosec}A}=2+\frac{\sin A}{\cot A-\text{cosec}A}$

Hence, proved.