If ƒ(x) = √(2tan(x)),then f-1(√(2)) =
Correct Answer :- b
Explanation : ƒ(x) = √(2tan(x))
x = √(2tan(x))
x2 = (2tan(x))
(x2)/2 = tan(x)
tan-1(x^2)/2 = x
By putting x = √2, we get
x = 1
The value of cos15º− sin15º is
If 2tan−1(cos x) = tan−1(2cosec x) , then x =
If 2 tan-1 (cos x) = tan -1(2 cosec x),
2tan-1(cos x) = tan-1 (2 cosec x)
= tan-1(2 cosec x)
= cot x cosec x = cosec x = x = π/4
tan−1(−2) + tan−1(−3) is equal to
tan-1(-2) + tan-1(-3)
The values of x which satisfy the trigonometric equation are :
The maximum value of sin x + cos x is
The value of tan150 + cot150 is
The value of tan150 + cot 150
is equal to
The number of solutions of the equation sin-1 x - cos-1 x = sin-1(1/2) is
Hence , the given equation has only one solution.
cot-1a - cot-1b + cot-1 b - cot-1 c - cot-1 a = 0
What is the maximum and minimum value of sin x +cos x?
Let y= sin x + cos x
dy/dx=cos x- sin x
For maximum or minimum dy/dx=0
Setting cosx- sin x=0
We get cos x = sin x
x= π/4, 5π/4———-
Whether these correspond to maximum or minimum, can be found from the sign of second derivative.
d^2y/dx^2=-sin x - cos x=-1/√2–1/√2 (for x=π/4) which is negative. Hence x=π/4 corresponds to maximum.For x=5π/4
d^2y/dx^2=-(-1/√2)-(-1/√2)=2/√2 a positive quantity. Hence 5π/4 corresponds to minimum
Maximum value of the function
y= sin π/4 + cos π/4= 2/√2=√2
Minimum value is
Sin(5π/4)+cos (5π/4)=-2/√2=-√2
sin (200)0 + cos (200)0 is
Because both sin 2000 and cos 2000 lies in 3rd quadrant. In 3rd quadrant values of sin and cos are negative.
If cos(-1)x + cos(-1)y = 2π, then the value of sin(-1)x + sin(-1)y is
If cos(-1)x + cos(-1) y = 2π, then the value of sin(-1)x + sin(-1)y = π−2π = −π.
Domain of f(x) = sin−1x−sec−1x is
Since sin−1x is defined for |x|⩽1, and sec−1x is defined for |x|⩾1,therefore,f(x) is defined only when|x|=1.so, Df = {−1,1}.
The value of sin is
Put therefore the given expressionis sin2θ = 2sinθcosθ
If 5 sin θ = 3, then is equal to
= [(5/4) + (3/4)] / [(5/4) - (3/4)]
=(8/4) / (2/4)
=4
If sin A + cos A = 1, then sin 2A is equal to
(sinA+cosA)2
= sin2A+cos2A+2sinAcosA
=11+sin2A=1sin2A=0.
(because Sin 2A = 2sin A cos A)
If θ = cos-1, then tan θ is equal to
Therefore, tanθ =
The number of solutions of the equation cos-1(1-x) - 2cos-1 x = π/2 is
As no value of x in (0, 1) can satisfy the given equation.Thus, the given equation has only one solution.
tan (sin−1x) is equal to
If x ∈ R, x ≠ 0, then the value of sec θ + tan θ is
The value of cos 1050 is
If cos(2sin−1x) = 1/9 then x =
Put
sin-1 x = θ ⇒ x = sin θ
cot (cos−1x) is equal to
Put,
cos-1x = θ ⇒ x = cos θ ⇒ cos θ
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