JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

\sin ^{-1}(\sin x)+\cos ^{-1}(\cos y)+\sec ^{-1}(\sec z)=\frac{5 \pi}{2}

then the value of

\sin x+\cos y+\sec z

View answer and explanation

Correct option: B

-1

Option B (-1) is correct because the given equation is only possible if sin x = 1, cos y = -1, and sec z = -1, leading to sin x + cos y + sec z = -1.

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x

{x^2} + {y^2} + {z^2} = {r^2},

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