Question

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

Answer 1

Given that, $A+B={90}^{\circ }$

$⟹B={90}^{\circ }-A$

$\tan B=\tan ({90}^{\circ }-A)=\cot A$

$\sec B=\sec ({90}^{\circ }-A)=\csc A$

$\sin B=\sin ({90}^{\circ }-A)=\cos A$

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

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

$=\sqrt{\frac{1+1}{1}-1}$

$=\sqrt{2-1}=\sqrt{1}=1$