Question

Solve:
$(\csc A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$

Answer 1

$\begin{array}{c}\left(\cos ecA-\sin A\right)\left(\sec A-\cos A\right)=\frac{1}{\tan A+\cot A}\\ \left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)\\ \left(1-\frac{{\sin }^{2}A}{\sin }\right)\left(\frac{1-{\cos }^2}{\cos }A\right)\\ =\left(\frac{{\cos }^{2}A\times {\sin }^{2}A}{\sin A.\cos A}\right)=\cos A.\sin A\\ =\frac{\cos A.\sin A}{1}=\frac{\cos A\sin A}{{\sin }^{2}A+{\cos }^{2}A}\\ =\frac{1}{\frac{\sin A}{\cos A.\sin A}+\frac{{\cos }^{2}A}{\sin A.\cos A}}=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}\\ =\frac{1}{\tan A+\cot A}\end{array}$