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Correct Answer : b
Explanation : ƒ(x) = √(2tan(x))
x = √(2tan(x))
x^{2} = (2tan(x))
(x^{2})/2 = tan(x)
tan^{1}(x^2)/2 = x
By putting x = √2, we get
x = 1
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)
The values of x which satisfy the trigonometric equation are :
The value of tan15^{0} + cot 15^{0}
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^{1}a  cot^{1}b + 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
Because both sin 200^{0} and cos 200^{0} 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.
Since sin^{−1}x is defined for x⩽1, and sec^{−1}x is defined for x⩾1,therefore,f(x) is defined only whenx=1.so, D_{f} = {−1,1}.
Put therefore the given expressionis sin2θ = 2sinθcosθ
= [(5/4) + (3/4)] / [(5/4)  (3/4)]
=(8/4) / (2/4)
=4
(sinA+cosA)^{2}
= sin^{2}A+cos^{2}A+2sinAcosA
=11+sin^{2}A=1sin^{2}A=0.
(because Sin 2A = 2sin A cos A)
Therefore, tanθ =
The number of solutions of the equation cos^{1}(1x)  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.
If x ∈ R, x ≠ 0, then the value of sec θ + tan θ is
Put
sin^{1 }x = θ ⇒ x = sin θ
Put,
cos^{1}x = θ ⇒ x = cos θ ⇒ cos θ
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