cos2θ is not equal to
cos 2θ is equal to cos2θ - sin2θ = 2cos2θ - 1
is equal to
sin-1√3/5 = A
Sin A = √3/5 , cos A = √22/5
Therefore Cos-1√3/5 = B
Cos B = √3/5 , sin B = √22/5
sin(A+B) = sinA cosB + cosA sinB
= √3/5 * √3/5 + √22/5 * √22/5
= 3/25 * 22/25
= 25/25
= 1
2cos−1x = cos−1(2x2−1)holds true for all
This is true for all real values of x ∈ [0,1].
none of these : 7/25
If sin A + cos A = 1, then sin 2A is equal to
(Sin A + Cos A)2 = sin2A + cos2A + 2 sinAcosA
1 = 1 + Sin 2A
so, Sin 2A = 0
Hence A = 0
When x = π/2, then tan x, is
tan π/2 = n.d.i.e. not defined.
The value of the expression sinθ+cosθ lies between
Minimum value
and maximum value =
cos−1(cosx) = x is satisfied by ,
cos−1(cosx) = x if,
0⩽x⩽ π i.e. if , x∈ [0,π]
If ƒ(x) = tan(x), then f-1(1/√(3)) =
Let F(x) = tanx = y
f(-1) = tany = x
f(-1)(1/√3), tany = 1/√3
y = tan-(1/√3)
=π/6
tan x is periodic with period
The values of tan x repeats after an interval of π.
is equal to
The correct option is B.
The period of the function f(x) = tan 3x is
f(x) = tan 3x, as the period of tan x = π. 3x = x = π/3.
The range of tan−1 x is
The range of tan−1x is given by -π/2 to π/2
The value of is (a, b > 0)
The solution of the equation
But , x= -1/2 does not satisfy the given equation.
is equal to
= tan-1 a - tan-1 b + tan-1c + tan-1c - tan-1a
sin2250+sin2650 is equal to
sin2250 + sin2650 = cos2650 + sin2650 = 1
If θ = tan−1x then sin 2θ is equal to
As we know that : sin 2θ = , now if θ = tan-1x ⇒ x = tan θ ⇒ sin 2θ =
if x > 0, then tan-1x + tan-1(1/x) is equal to
If and x + y + z = xyz, then a value of tan−1x+tan−1y+tan−1z is
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