Question

Prove: $(\text{cosec}A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$

Answer 1

To prove $(\text{cosec}A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$
Lets take LHS and prove it equal to RHS,

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

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

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

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

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

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

$=\frac{1}{\tan A+\cot A}=RHS$
$\therefore (\text{cosec}A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$