1
JEE Main 2022 (Online) 27th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$${\sin ^1}\left( {\sin {{2\pi } \over 3}} \right) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} \right) + {\tan ^{ - 1}}\left( {\tan {{3\pi } \over 4}} \right)$$ is equal to :

A
$${{11\pi } \over {12}}$$
B
$${{17\pi } \over {12}}$$
C
$${{31\pi } \over {12}}$$
D
$$-$$$${{3\pi } \over {4}}$$
2
JEE Main 2022 (Online) 26th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If the inverse trigonometric functions take principal values then

$${\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right)$$ is equal to :

A
0
B
$${\pi \over 4}$$
C
$${\pi \over 3}$$
D
$${\pi \over 6}$$
3
JEE Main 2022 (Online) 25th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The value of $${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right)$$ is equal to :

A
$$ - {\pi \over 4}$$
B
$$ - {\pi \over 8}$$
C
$$ - {{5\pi } \over {12}}$$
D
$$ - {{4\pi } \over 9}$$
4
JEE Main 2022 (Online) 24th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$x * y = {x^2} + {y^3}$$ and $$(x * 1) * 1 = x * (1 * 1)$$.

Then a value of $$2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$$ is :

A
$${\pi \over 4}$$
B
$${\pi \over 3}$$
C
$${\pi \over 2}$$
D
$${\pi \over 6}$$
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