Question

Prove that $\frac{\tan A}{1-cotA}+\frac{cotA}{1-\tan A}=1+\tan A+cotA$ .

Answer 1

Now by taking L.H.S

$=\frac{\tan A}{1-cotA}+\frac{cotA}{1-\tan A}$

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

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

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

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

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

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

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

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

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

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

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

$=tanA+1+\mathrm{co}tA$

$=1+tanA+\mathrm{co}tA$

$=R.H.S$

$\Rightarrow L.H.S=R.H.S$

Hence proved