# If A + B = 90°, then tan A tan B+tan A cot Bsin A sec B−sin2 Bcos2 A is equal to

Question:

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

(a) $\cot ^{2} \mathrm{~A}$

(b) $\cot ^{2} \mathrm{~B}$

(c) $-\tan ^{2} A$

(d) $-\cot ^{2} A$

Solution:

We have:

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

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

We have to find the value of the following expression

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

So

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

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

$=\frac{\tan A \cot A+\tan A \tan A}{\sin A \operatorname{cosec} A}-\frac{\cos ^{2} A}{\cos ^{2} A}$

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

$=\tan ^{2} A$

$=\tan ^{2}\left(90^{\circ}-B\right)$

$=\cot ^{2} B$

Hence the correct option is $(b)$