Question

If $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}=1+secA.cosecA.$

Answer 1

${\sin }^{3}A-{\cos }^{3}A=(\sin A-\cos A)({\sin }^{2}A+{\cos }^{2}A+\sin A\cos A)=(\sin A-\cos A)(1+\sin A\cos A)$

$LHS=\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$

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

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

$=\frac{1}{\sin A-\cos A}(\frac{{\sin }^{3}A-{\cos }^{3}A}{\sin A\cos A})$

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

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

$=1+\text{sec}A\text{cosec}A=RHS$

Hence Proved