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Question 139
If $$\alpha + \beta = 90^\circ$$ then the expression $$\frac{\ \tan\ \alpha\ }{\tan\ \beta\ }+\sin^2\alpha\ +\sin^2\beta\ $$ is equal to :
Solution
Given  $$\ \alpha\ +\beta\ \ =90$$  $$=$$>  $$\ \alpha\ =90-\beta\ $$
$$\frac{\ \tan\ \alpha\ }{\tan\ \beta\ }+\sin^2\alpha\ +\sin^2\beta\ =\frac{\ \tan\ \alpha\ }{\tan\ \left(90-\alpha\ \right)\ }+\sin^2\alpha\ +\sin^2\left(90-\alpha\ \right)=\frac{\ \tan\ \alpha\ }{\cot\ \alpha\ }+\sin^2\alpha\ +\cos^2\alpha=\tan^2\alpha\ +1=\sec^2\alpha\ $$
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