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UPSC CDS Math Trigonometric Ratios and Trigonometric Identities Questions

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Question : 10 of 107

Marks: +1, -0

What is the value of

\sin^2

\sin^2

0
1
2
4

Solution:

\sin^{2}25^{\circ}+\sin^{2}65^{\circ}

=\sin^{2}25^{\circ}+\sin^2(90^{\circ}-25^{\circ})

=\sin^{2}25^{\circ}+\cos^{2}25^{\circ}=160

(c) We know that the maximum value of sin θ = 1

And the maximum value of cos θ = 1

Therefore,

\sec\alpha = \frac{1}{4}

and

\csc\gamma = \frac{1}{2}

are not possible

tan θ can assume any value. Therefore, only tan β = 20 is possible.

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