Question

$(cosecA-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$
[Hint: Simplify LHS and RHS separately]

Answer 1

L.H.S$=(\csc A-\sin A)(\sec A-\cos A)$

$=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)$ where $\csc A=\frac{1}{\sin A}$ and $\sec A=\frac{1}{\cos A}$

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

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

$=\sin A\cos A$

$R.H.S=\frac{1}{\tan A+\cot A}$

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

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

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

Hence L.H.S$=$R.H.S

Proved