Question

Prove that $\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{cosecA+\cot A-1}=1$

Answer 1

Consider LHS $=\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{cosecA+\cot A-1}$

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

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

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

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

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

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

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

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

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

$=1$

$=RHS$

Hence proved