1
JEE Main 2017 (Online) 9th April Morning Slot
+4
-1
A value of x satisfying the equation sin[cot−1 (1+ x)] = cos [tan−1 x], is :
A
$$- {1 \over 2}$$
B
$$-$$ 1
C
0
D
$${1 \over 2}$$
2
JEE Main 2017 (Online) 8th April Morning Slot
+4
-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 :
A
$${\pi \over 4} + {1 \over 2}{\cos ^{ - 1}}\,{x^2}$$
B
$${\pi \over 4} + {\cos ^{ - 1}}\,{x^2}$$
C
$${\pi \over 4} - {1 \over 2}{\cos ^{ - 1}}\,{x^2}$$
D
$${\pi \over 4} - {\cos ^{ - 1}}\,{x^2}$$
3
JEE Main 2015 (Offline)
+4
-1
Let $${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right),$$
where $$\left| x \right| < {1 \over {\sqrt 3 }}.$$ Then a value of $$y$$ is :
A
$${{3x - {x^3}} \over {1 + 3{x^2}}}$$
B
$${{3x + {x^3}} \over {1 + 3{x^2}}}$$
C
$${{3x - {x^3}} \over {1 - 3{x^2}}}$$
D
$${{3x + {x^3}} \over {1 - 3{x^2}}}$$
4
JEE Main 2013 (Offline)
+4
-1
If $$x, y, z$$ are in A.P. and $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$$ and $${\tan ^{ - 1}}z$$ are also in A.P., then :
A
$$x=y=z$$
B
$$2x=3y=6z$$
C
$$6x=3y=2z$$
D
$$6x=4y=3z$$
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