Question

Show that:
$\frac{\sin A}{\sec A+\tan A-1}+\frac{\tan A}{\csc A+\cot A-1}=1$

Answer 1

$\begin{array}{c}Thereshouldbe\cos Ainplaceof\tan A-1\sec ondgterm\\ LHS=\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\cos ecA+cotA}\\ \Rightarrow \frac{\sin A}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}-1}+\frac{\cos A}{\frac{1}{\sin A}+\frac{\cos A}{\sin A}+1}\\ \Rightarrow \frac{\cos A\sin A}{1+\sin A-\cos A}+\frac{\cos A\sin A}{1+\cos A-\sin A}\\ \Rightarrow \cos A\sin A\left[\frac{1}{1+\sin A\cos A}+\frac{1}{1+\cos A-\sin A}\right]\\ \Rightarrow \cos A\sin A\left[\frac{1+\cos A-\sin A+1+\sin A-\cos A}{\left(1+\sin A-\cos A\right)\left(1+\cos A-siNA\right)}\right]\\ \Rightarrow \frac{2\cos A\sin A}{1-\left({\cos }^{2}A+{\sin }^{2}A-2\cos A\sin A\right)}=\\ \Rightarrow \frac{2\cos A\sin A}{1-\left(1-2\cos A\sin A\right)}=1=RHS.\end{array}$