Question

If $A+B={90}^{\circ }$ then prove that $\sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}-\frac{{\sin }^{2}B}{{\cos }^{2}A}}=\tan A$

Answer 1

Here $A+B={90}^{\circ }$

$\Rightarrow B={90}^{\circ }-A$

$=\sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}-\frac{{\sin }^{2}B}{{\cos }^{2}A}}$

$=\sqrt{\frac{\tan A\tan ({90}^{\circ }-A)+\tan A\cot ({90}^{\circ }-A)}{\sin A\sec ({90}^{\circ }-A)}-\frac{{\sin }^2({90}^{\circ }-A)}{{\cos }^{2}A}}$

$=\sqrt{\frac{\tan A\cot A+\tan A\tan A}{\sin AcosecA}-\frac{{\cos }^{2}A}{{\cos }^{2}A}}$

$=\sqrt{\frac{\tan A\cot A+{\tan }^{2}A}{\sin AcosecA}-1}$

$=\sqrt{\frac{1+{\tan }^{2}A}{1}-1}$

$=\sqrt{1+{\tan }^{2}A-1}$

$=\sqrt{{\tan }^{2}A}$

$=\tan A$

Hence proved.