JEE Main 2019 (Online) 9th January Evening Slot | Inverse Trigonometric Functions Question 66 | Mathematics

If $${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$ (x > $$3 \over 4$$), then x is equal to :

$${{\sqrt {145} } \over {10}}$$

$${{\sqrt {145} } \over {11}}$$

$${{\sqrt {145} } \over {12}}$$

$${{\sqrt {146} } \over {12}}$$

JEE Main 2017 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

A value of x satisfying the equation sin[cot⁻¹ (1+ x)] = cos [tan⁻¹x], is :

$$ - {1 \over 2}$$

$$-$$ 1

$$ {1 \over 2}$$

JEE Main 2017 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

The value of tan^-1 $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right],$$ $$\left| x \right| < {1 \over 2},x \ne 0,$$ is equal to :

$${\pi \over 4} + {1 \over 2}{\cos ^{ - 1}}\,{x^2}$$

$${\pi \over 4} + {\cos ^{ - 1}}\,{x^2}$$

$${\pi \over 4} - {1 \over 2}{\cos ^{ - 1}}\,{x^2}$$

$${\pi \over 4} - {\cos ^{ - 1}}\,{x^2}$$

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