Question

Prove that ${\cot }^{2}A\left(\frac{\sec A-1}{1+\sin A}\right)+{\sec }^{2}A\left(\frac{\sin A-1}{1+\sec A}\right)=0$.

Answer 1

${\cot }^{2}A\left(\frac{\sec A-1}{1+\sin A}\right)+{\sec }^{2}A\left(\frac{\sin A-1}{1+\sec A}\right)=0$

LHS$=\frac{{\cos }^{2}A}{{\sin }^{2}A}\left(\frac{\frac{1}{cosA}-1}{1+\sin A}\right)+\frac{1}{{\cos }^{2}A}\left(\frac{\sin A-1}{1+\frac{1}{\cos A}}\right)$

$=\frac{{\cos }^{2}A}{{\sin }^{2}A}\left(\frac{1-\cos A}{\cos A(1+\sin A)}\right)+\frac{1}{{\cos }^{2}A}\left(\frac{\cos A(\sin A-1)}{\cos A+1}\right)$

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

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

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

$=\frac{1}{1+\cos A}\left[\frac{{\cos }^{2}A+{\sin }^{2}A-1}{(1+\sin A)\cos A}\right]$

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

$=\frac{1}{\cos A(1+\cos A)(1+\sin A)}\left(0\right)=0$

Hence proved.