Question

What is the value of $\sec ({90}^0-A)-\cot A\cos ({90}^0-A)\tan ({90}^0-A)$ ?

Answer 1

$\sec ({90}^0-A)-\cot A\cos ({90}^0-A)\tan ({90}^0-A)=co\sec A-\cot A\cos ({90}^0-A)\tan ({90}^0-A)$
$=\frac{1}{\sin A}-\frac{\cos A}{\sin A}\times \sin A\times \frac{\cos A}{\sin A}$
$=\frac{1-{\cos }^{2}A}{\sin A}=\frac{{\sin }^{2}A}{\sin A}=\sin A$