Question : If $A+B=90^{\circ}$, then the expression $\frac{\cot A}{\cot B}+\cos ^2 A+\cos ^2 B$ is equal to:
Option 1: $\cot ^2 B$
Option 2: $\operatorname{cosec}^2 A$
Option 3: $\cot ^2 A$
Option 4: $\operatorname{cosec}^2 B$
Correct Answer: $\operatorname{cosec}^2 A$
Solution :
$A+B=90^{\circ}$
$B=90^{\circ}-A$
$\frac{\cot A}{\cot B}+\cos ^2 A+\cos ^2 B$
= $\frac{\cot A}{\cot (90^{\circ}-A)}+\cos ^2 A+\cos ^2 (90^{\circ}-A)$
= $\frac{\cot A}{\tan A}+\cos ^2 A+\sin ^2A$
= $\cot^2 A +1$
= $\operatorname{cosec}^2 A$
Hence,the correct answer is $\operatorname{cosec}^2 A$.
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