If Sec-1(x/a)- Sec-1(x/b)= Sec-1 b- Sec-1a then x=
ab
b/a
a/b
1/ab
Let Then which one of the following is true
I < 2/3 and J > 2
I < 2/3 and J < 2
I > 2/3 and J > 2
The ascending order of A= Sin-1(sin 8π/7),B= Cos-1(cos 8π/7), ), C=Tan-1(tan 8π/7) is
B,A, C
B,C,A
A,B,C
A,C,B
If sin-1 (3/5)+sin-1(5/13)= sin-1 x, then x=
51/65
52/65
56/65
none of these
sin-1(2cos2 x-1)cos-1(1-2sin2 x)
π/2
π/3
π/4
π/6
Sin-1(24/25) +Tan-1(5/12) =
Tan-1(27/11)
Tan-1(16/63)
Sin-1(16/65)
Cos-1(-36/325)
4 Tan-1 1/5- Tan-1 1/239 =
π
3π/4
Sec-1√34/5+ Cosec-1√17 =
If Cos-1 x= Tan-1 x, then sin(Cos-1 x)=
x
x2
1/x
1/ x2
The domain of Sinh-1 2x is
R
[0, ∞)
[1/2, ∞)
(-1/2, 1/2)
4 Tan-1 1/5- Tan-1 1/70+Tan-1 1/99 =
If Tan-1 (sec x + tan x)=π/4+kx then k=
2
4
1/2
1/4
The range of f(x)= Sin-1x-cos-1x + Tan-1x is
(0, π)
(π/4, 3π/4)
(-π/4, π/4)
[0, 3π/4]
tan (1/2 cos-1(0)) =
0
-1
1
If Tan-1(x-1/x-2)+ Cot-1 (x+2/x+1)=π/4, then x=
1/√2
±1/√2
±1/√3
1/√3
Sin-1(3/5)+Sin-1(8/17)=
Sin-1(56/65)
Sin-1(33/65)
Sin-1(77/85)
Sin-1(3/5)
Tan (tan-1 1/2+ tan-11/3) =
5
Sin-1(-√2/2) + Cos-1(-1/2)-Tan-1(-√3)-Cot-1(1/√3) =
5π/6
5π/12
7π/12
7π/6
Sin (2Tan-13/4) =
Sin(4Tan-11/3)
Sin(2 Tan-11/3)
Cos(2 Tan-11/7)
Cos(4 Tan-11/7)
If Tan-1 x+ Tan-1 y+ Tan-1 z=π, then x+y+z=
xyz
2xyz
If Sin-1 x+ Sin-1 y+ Sin-1 z=π then x2+y2+z2+2xyz=
Tan-1 3/2-Tan-11/5 =
2 Tan-1 1/3+Tan-1 1/7=
The domain of Cos-1 (2/2+sinx) in [0,2π] is
[0,π)
[0, π/2]
[-2,-1] U[1,2]
none
The domain of f(x)=Tan-1 √x(x+3) + sin-1√x2+3x+1 is
(-3, 0)
(3, 0)
{-3, 0}
{0, 3}
Sin-1(sin 2π/3) =
π/12
Tan (π/4+1/2cos-1a/b) +tan (π/4-1/2cos-1a/b) =
2a/b
2b/a
Cot-1(4/3)-Cos-1(15/8) =
Cot-1(16/65)
Cot-1(84/65)
Cot-1(84/85)
Cot-1(84/13
If Cosec-1 x= 2Cot-1 7+ Cos-1(3/4) then the value of x is
44/117
125/117
24/7
5/3
Sin-1 (16/65) +2 Tan-1 (1/5) =
Cos-1(3/4)
Cos-1(4/3)
Cos-1(5/4)
Cos-1(4/5)