JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

A continuous even periodic function

f

with period 8 is such that

f(0)=0, f(1)=-2, f(2)=1, f(3)=2, f(4)=3

then the value of

\tan ^{-1}\left(\tan \left(f(-5)+f(20)+\cos ^{-1}(f(-10)+f(17))\right)\right)

is equal to:

View answer and explanation

Correct option: B

5-2 \pi

Option B (

5-2 \pi

) is correct because after simplification using the properties of trigonometric functions and inverse trigonometric functions, the expression evaluates to

5-2 \pi

More Questions from INVERSE TRIGONOMETRY

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The value of

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

is- The value of

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

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

is Find

\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

, then

x

is equal to If

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

then

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

is equal to

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

is equal to