WB JEE 2009 | Inverse Trigonometric Functions Question 2 | Mathematics

WB JEE 2009

MCQ (Single Correct Answer)

-0.25

$$\tan \left[ {{\pi \over 4} + {1 \over 2}{{\cos }^{ - 1}}\left( {{a \over b}} \right)} \right] + \tan \left[ {{\pi \over 4} - {1 \over 2}{{\cos }^{ - 1}}\left( {{a \over b}} \right)} \right]$$ is equal to

$${{2a} \over b}$$

$${{2b} \over a}$$

$${a \over b}$$

$${b \over a}$$

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)

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