JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Hard

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Analyzing +4 | -1

Domain of the function

f(x)=2 \operatorname{Sin}^{-1} \sqrt{1-x}+\operatorname{Sin}^{-1}[2 \sqrt{x(1-x)}]

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Correct option: A

0 \leq x \leq 1

Option A (

0 \leq x \leq 1

) is correct because it satisfies the domain restrictions of both inverse sine functions, ensuring that the arguments are within [-1, 1].

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\sin \left(2 \tan ^{-1}(1 / 3)\right)+

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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)