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