WB JEE 2010 | Inverse Trigonometric Functions Question 3 | Mathematics

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WB JEE 2010

MCQ (Single Correct Answer)

-0.25

Value of $${\tan ^{ - 1}}\left( {{{\sin 2 - 1} \over {\cos 2}}} \right)$$ is

$${\pi \over 2} - 1$$

$$1 - {\pi \over 4}$$

$$2 - {\pi \over 2}$$

$${\pi \over 4} - 1$$

WB JEE 2011

MCQ (Single Correct Answer)

-0.25

The solution set of the inequation $${\cos ^{ - 1}}x < {\sin ^{ - 1}}x$$ is

[$$-$$1, 1]

$$\left[ {{1 \over {\sqrt 2 }},1} \right]$$

[0, 1]

$$\left( {{1 \over {\sqrt 2 }},1} \right]$$

WB JEE 2021

MCQ (Single Correct Answer)

-0.25

For $$y = {\sin ^{ - 1}}\left\{ {{{5x + 12\sqrt {1 - {x^2}} } \over {13}}} \right\};\left| x \right| \le 1$$, if $$a(1 - {x^2}){y_2} + bx{y_1} = 0$$ then (a, b) =

(2, 1)

(1, $$-$$1)

($$-$$1, 1)

(1, 2)

WB JEE 2018

MCQ (Single Correct Answer)

-0.25

If $$0 \le A \le {\pi \over 4}$$, then $${\tan ^{ - 1}}\left( {{1 \over 2}\tan 2A} \right) + {\tan ^{ - 1}}(\cot A) + {\tan ^{ - 1}}({\cot ^3}A)$$

$${\pi \over 4}$$

$$\pi$$

$${\pi \over 2}$$

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