Question

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

Answer 1

L.H.S

$(cosecA-\sin A)(\sec A-\cos A)$

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

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

$\frac{{\cos }^{2}A\cdot {\sin }^{2}A}{\sin A\cos A}$

$=\sin A\cos A$

R.H.S.

$=\frac{1}{\tan A+\cot A}$

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

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

$=\sin A\cos A$

Thus, L.H.S$=$R.H.S.