Question

If $A+B={90}^0$, The value of $\sqrt{\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}-\frac{{\sin }^{2}B}{{\cos }^{2}A}}=$

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

Answer 1

The correct option is A $\tan A$

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

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

$[A+B=90]$

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

as $\left[\begin{array}{cc}\begin{array}{c}\tan (90-A)=\cot A\\ \cot (90-\theta )=\tan \theta \end{array}& \begin{array}{c}\sec (90-\theta )=\csc \theta \\ \sin (90-\theta )=\cos \theta \\ \cos (90-\theta )=\cos \theta \end{array}\end{array}\right]$

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

as $\cot A=\frac{1}{\tan A}$

$\csc A=\frac{1}{\sin A}$

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

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