Question

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

• A. $\tan A$
• B. $\sin A$
• C. $\cot A$
• D. $\csc A$

Answer 1

Answer: (A)

$\tan A$

If $A+B={90}^{\circ }$ then

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

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

$\tan B=\cot (90-B)=\cot A$
$\cot B=\tan (90-B)=\tan A$
$\cos B=\sin (90-B)=\sin A$

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

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