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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

If

{\sec ^{ - 1}}\sqrt {1 + {x^2}} + \mathrm{cose}{\mathrm{c}^{ - 1}}\frac{{\sqrt {1 + {y^2}} }}{y} + {\cot ^{ - 1}}\frac{1}{z} = \pi

, then

x + y + z

is equal to

View answer and explanation

Correct option: A

xyz

Option A (

xyz

) is correct because the given equation can be transformed into

{\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi

, which implies

x + y + z = xyz

.

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\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

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\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

x

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

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

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