If cotcos-1x = sectan-1ab2 - a2, then x is equal to
b2b2 - a2
a2b2 - a2
2b2 - a2a
2b2 - a2b
Number of solutions ofthe equation
tan-112x + 1 + tan-114x + 1 = tan-12x2 is
1
2
3
4
The value of tan-112 + tan-113 + tan-178 is
tan-178
cot-115
tan-115
tan-12524
C.
tan-112 + tan-113 + tan-178= tan-112 + 131 - 12 . 13 + tan-178 ∵ tan-1a + tan-1b = tan-1a + b1 - ab= tan-15656 + tan-178= tan-11 + tan-178= tan-11 + 781 - 78 = tan-115
If tan-1x - 1 + tan-1x + tan-1x + 1 = tan-13x, then x is
± 12
0, 12
0, - 12
0, ± 12
The domain of the function sin-1log2x22 is
[- 1, 2] - {0}
[- 2, 2] - (- 1, 1)
[- 2, 2] - {0}
[1, 2]
3tan-1a is equal to
tan-13a + a31 + 3a2
tan-13a - a31 + 3a2
tan-13a + a31 - 3a2
tan-13a - a31 - 3a2
The value of sinsin-113 + sec-13 + costan-112 + tan-12 is :
The value of cot-19 + csc-1414 is given by :
0
π4
π2
tan-12
cos-112 + 2sin-112 is equal to :
π6
π3
2π3
A particle possess two velocities simultaneously at an angle of tan-1125; to each other. Their resultant is 15 m/s. If one velocity is 13 m/s, then the other will be :
5 m/s
4 m/s
12 m/s
13 m/s