If $\cos ^{-1}(x / a)+\cos ^{-1}(y / b)=\alpha$. Then $x^2 / a^2+y^2 / b^2$ is equal to

$(2 x y / a b) \sin \alpha+\cos ^2 \alpha$

$(2 x y / a b) \cos ^2 \alpha+\sin \alpha$

$(2 x y / a b) \sin ^2 \alpha+\cos \alpha$

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

\cos ^{-1}(x / a)+\cos ^{-1}(y / b)=\alpha

. Then

x^2 / a^2+y^2 / b^2

is equal to

View answer and explanation

Correct option: A

(2 x y / a b) \cos \alpha+\sin ^2 \alpha

The correct answer is A (

(2 x y / a b) \cos \alpha+\sin ^2 \alpha

) because using the cosine addition formula and algebraic manipulation, we can arrive at this expression.

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2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)