Question

Show that $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=1+\sec AcosecA$

Answer 1

L.H.S

$\Rightarrow \frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$

$\Rightarrow \frac{\frac{\sin A}{\cos A}}{\frac{\sin A-\cos A}{\sin A}}+\frac{\frac{\cos A}{\sin A}}{\frac{\cos A-\sin A}{\cos A}}$

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

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

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

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

$\Rightarrow \frac{(1+\cos A\sin A)}{\cos A\sin A}$

$\Rightarrow 1+\sec A\csc A$

Hence, proved.