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Inverse Trigonometric Functions Test - 3

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Inverse Trigonometric Functions Test - 3
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  • Question 1
    2 / -0.83

    If ƒ(x) = √(2tan(x)),then f-1 (√(2)) =

    Solution

    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

  • Question 2
    2 / -0.83

    The value of cos15º −sin15º  is

    Solution

      

     

  • Question 3
    2 / -0.83

    If  2tan−1 (cos x) = tan−1 (2cosec x) , then x =

    Solution

    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

  • Question 4
    2 / -0.83

    tan−1 (−2) + tan−1 (−3) is equal to  

    Solution

    tan-1 (-2) + tan-1 (-3)

     

  • Question 5
    2 / -0.83

    The values of x which satisfy the trigonometric equation   are :

    Solution



  • Question 6
    2 / -0.83

    The maximum value of  sin x  + cos x is

    Solution

  • Question 7
    2 / -0.83

    The value of  tan150   + cot150   is

    Solution

    The value of tan150  + cot 150  

  • Question 8
    2 / -0.83

     is equal to  

    Solution

  • Question 9
    2 / -0.83

    The number of solutions of the equation  sin-1  x - cos-1  x = sin-1 (1/2) is

    Solution




    Hence , the given equation has only one solution.

  • Question 10
    2 / -0.83

    Solution

    cot-1 a - cot-1 b + cot-1  b - cot-1  c - cot-1   a = 0

  • Question 11
    2 / -0.83

    What is the maximum and minimum value of sin x +cos x?

    Solution

    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, 5 π/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=5 π/4

    d^2y/dx^2=-(-1/√2)-(-1/√2)=2/√2 a positive quantity. Hence 5 π/4 corresponds to minimum

    Maximum value of the function

    y= sin π/4 + cos π/4= 2/√2=√2

    Minimum value is

    Sin(5 π/4)+cos (5 π/4)=-2/√2=-√2

  • Question 12
    2 / -0.83

    sin (200)0   + cos (200)0 is

    Solution

    Because both  sin 2000  and cos 2000  lies in 3rd quadrant. In 3rd quadrant values of sin and cos are negative.

  • Question 13
    2 / -0.83

    If cos(-1) x + cos(-1) y = 2 π, then the value of sin(-1) x + sin(-1) y is   

    Solution

    If cos(-1) x + cos(-1)  y = 2 π, then the value of sin(-1) x + sin(-1) y  = π−2 π= −π.

  • Question 14
    2 / -0.83

    Domain of f(x) = sin−1 x −sec−1 x is

    Solution

    Since  sin−1 x  is defined for  |x|⩽1, and sec−1 x is defined for  |x|⩾1,therefore,f(x) is defined only when|x|=1.so, Df  = {−1,1}.

  • Question 15
    2 / -0.83

    The value of sin   is  

    Solution

    Put    therefore the given expressionis  sin2 θ= 2sin θcos θ

  • Question 16
    2 / -0.83

    If  5 sin θ= 3, then   is equal to

    Solution


    = [(5/4) + (3/4)] / [(5/4) - (3/4)]
    =(8/4) / (2/4)
    =4

  • Question 17
    2 / -0.83

    If sin  A  + cos  A  = 1, then sin 2A is equal to

    Solution

    (sinA+cosA)2
    = sin2 A+cos2 A+2sinAcosA
    =11+sin2 A=1sin2 A=0.
    (because Sin 2A = 2sin A cos A)

  • Question 18
    2 / -0.83

    If  θ= cos-1, then tan  θis equal to  

    Solution


    Therefore, tan θ=

  • Question 19
    2 / -0.83

    The number of solutions of the equation  cos-1 (1-x) - 2cos-1 x = π/2 is

    Solution

    As no value of x in (0, 1) can satisfy the given equation.Thus, the given equation has only one solution.

  • Question 20
    2 / -0.83

    tan  (sin−1 x) is equal to

    Solution

  • Question 21
    2 / -0.83

    If  x  ∈R, x  ≠0, then the value of sec  θ+ tan  θis  

    Solution

     

  • Question 22
    2 / -0.83

    The value of  cos 1050 is

    Solution

  • Question 23
    2 / -0.83

    If  cos(2sin−1 x) = 1/9  then x =

    Solution

    Put

    sin-1   x = θ⇒x = sin θ


  • Question 24
    2 / -0.83

    Solution






  • Question 25
    2 / -0.83

    cot  (cos−1 x) is equal to

    Solution

    Put,
    cos-1 x = θ⇒x = cos θ⇒cos θ

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