Trigonometric Function MCQ Quiz - Objective Question with Answer for Trigonometric Function - Download Free PDF

Last updated on May 3, 2025

Latest Trigonometric Function MCQ Objective Questions

Trigonometric Function Question 1:

Find dydx, if y = tan1[8x115x2]

  1. 51+25x231+9x2
  2. 51+25x2+31+9x2
  3. 81+25x2
  4. None of these
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 51+25x2+31+9x2

Trigonometric Function Question 1 Detailed Solution

Concept:

tan1x+tan1y=tan1[x+y1xy]

d(tan1x)dx=11+x2

Calculation:

Given: y = tan1[8x115x2]

y=tan1[5x+3x15x3x]

As we know that, tan1x+tan1y=tan1[x+y1xy]

So, y=tan1[5x+3x15x3x]= tan-1 5x + tan-1 3x

Differentiating with respect to x, we get

dydx=d(tan15x)dx+d(tan13x)dx

=51+(5x)2+31+(3x)2

=51+25x2+31+9x2

Trigonometric Function Question 2:

If y=sin2(cot11+x1x) then dydx is

  1. 12
  2. 2
  3. 12
  4. 2
  5. -5

Answer (Detailed Solution Below)

Option 3 : 12

Trigonometric Function Question 2 Detailed Solution

y=sin2(cot11+x1x)

Let, cotθ=1+x1x

cot2θ=1+x1x

x=cot2θ1cot2θ+1=cos2θsin2θcos2θ+sin2θ=cos2θ

Now,

y=sin2θ

dydx=dyd(θ)×d(θ)dx

dydx=2sinθcosθ×12sin(2θ)

dydx=12

Trigonometric Function Question 3:

If tan1(23x+1)=cot1(33x+1), then which one of the following is true ?

  1. There is no real value of x satisfying the above equation.
  2. There is one positive and one negative real value of x satisfying the above equation.
  3. There are two real positive values of x satisfying the above equation.
  4. There are two real negative values of x satisfying the above equation.

Answer (Detailed Solution Below)

Option 1 : There is no real value of x satisfying the above equation.

Trigonometric Function Question 3 Detailed Solution

Concept:

Inverse Tangent and Cotangent Relationships:

  • The inverse cotangent function can be written in terms of the inverse tangent function: cot-1 θ = (π/2) - tan-1 θ.
  • This relationship is useful in simplifying equations involving both tan-1 and cot-1.

 

Calculation:

Given the equation:

tan-1 (2 / (3x + 1)) = cot-1 (3 / (3x + 1))

We use the identity cot-1 θ = (π/2) - tan-1 θ to rewrite the equation as:

tan-1 (2 / (3x + 1)) = (π/2) - tan-1 (3 / (3x + 1))

Taking the tangent of both sides:

(2 / (3x + 1)) = (3 / (3x + 1))

This leads to the contradictory equation:

2 = 3

Therefore, there is no solution to this equation.

Conclusion:

The correct answer is:

  • Option (1): There is no real value of x satisfying the above equation.

Trigonometric Function Question 4:

If y=sin1x+y, then dydx= __________ (where x ∈ (0, 1)) 

  1. 1(2y1)1x2
  2. 1(12y)1x2
  3. 1(2y1)x21
  4. 1(2y+1)1x2

Answer (Detailed Solution Below)

Option 1 : 1(2y1)1x2

Trigonometric Function Question 4 Detailed Solution

Calculation

Given: y=sin1x+y

Squaring both sides:

y2=sin1x+y

Differentiating both sides with respect to x:

2ydydx=11x2+dydx

2ydydxdydx=11x2

dydx(2y1)=11x2

dydx=1(2y1)1x2 

Hence option 1 is correct.

Trigonometric Function Question 5:

If y=sinx+y then find dydx at x=0,y=1

  1. 0
  2. 1
  3. 2
  4. -1

Answer (Detailed Solution Below)

Option 2 : 1

Trigonometric Function Question 5 Detailed Solution

Calculation

y=sinx+y

y2=sinx+y

2ydydx=cosx+dydx

2ydydxdydx=cosx

dydx(2y1)=cosx

dydx=cosx2y1

Substitute x = 0 and y = 1:

dydx=cos(0)2(1)1

dydx=121

dydx=11

dydx=1

dydx=1 at x = 0, y = 1

Hence option 2 is correct

Top Trigonometric Function MCQ Objective Questions

If y = tan (cot−1 x), then  dydx at x = 1 is equal to

  1. 1
  2. -1
  3. π/4
  4. 0

Answer (Detailed Solution Below)

Option 2 : -1

Trigonometric Function Question 6 Detailed Solution

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Concept:

cot1x=tan1(1x)

tan (tan-1 x) = x

Calculation:

Given:

y = tan (cot−1 x)                                      

 y=tan[tan1(1x)]               

⇒ y = 1/x                                (∵tan (tan-1 x) = x)

Differentiating with respect to x, we get

dydx=1x2

At x = 1

dydx=11=1

Differentiate tan1[5+x15x] with respect to x.

  1. 1 + x2
  2. x1+x2
  3. 11x2
  4. 11+x2

Answer (Detailed Solution Below)

Option 4 : 11+x2

Trigonometric Function Question 7 Detailed Solution

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Concept:

  • tan1a+tan1b=tan1(a+b1ab)      ----(1)
  • da/dx = 0, where a is any constant
  • ddxtan1x=11+x2      ----(2)
  • ddx(a+b)=dadx+dbdx

Calculation:

ddxtan1[5+x15x]

ddx(tan15+tan1x)      [Using (1)]

ddx(tan15)+ddx(tan1x)

0+11+x2      [Using (2)]

11+x2

What is the value of

 sin33cos57+sec62sin28+cos33sin57+cosec62cos28tan15tan35tan60tan55tan75

  1. 2√3
  2. √3
  3. 2
  4. 33

Answer (Detailed Solution Below)

Option 2 : √3

Trigonometric Function Question 8 Detailed Solution

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Given:

? = sin33cos57+sec62sin28+cos33sin57+cosec62cos28tan15tan35tan60tan55tan75

Formula:

sin (90 - θ) = cos θ

tan (90 - θ) = cot θ

Calculation:

⇒ sin 33°. cos57° = sin(90° - 57°).cos57° = cos57°.cos57° = cos257°

⇒ sec 62°.sin28° = 1/cos 62° × sin 28° = sin (90° - 62°) × 1/cos 62°

= cos62° × 1/cos 62° = 1

⇒ cos33°. sin 57° = sin (90° - 33°) . sin 57° = sin257°

⇒ cosec 62°. cos 28° = cos 28° × 1/sin 62°

= cos 28° × 1/sin(90° - 28°)

= cos 28°/cos 28° = 1

⇒ tan 15°.tan 35°.tan 60°.tan 55°.tan 75° = (sin15°/cos15°) × (sin75°/cos75°) × (sin55°/cos55° ) × (sin35°/cos35°) × √3

= √3

Then,

⇒ ? = (cos257° + 1 + sin257° + 1)/√3

⇒ ? = 3/√3

⇒ ? = √3

sin33cos57+sec62sin28+cos33sin57+cosec62cos28tan15tan35tan60tan55tan75 = √3

Find d(sinxt)dt 

  1. t cos xt
  2. x cos xt
  3. -x cos xt
  4. t sin xt

Answer (Detailed Solution Below)

Option 2 : x cos xt

Trigonometric Function Question 9 Detailed Solution

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Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

  • Chain Rule: ddx[f(g(x))]=f(g(x))g(x)
  • Product Rule: ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)

 

Calculation:

We have to find the value of d(sinxt)dt

d(sinxt)dt=d(sinxt)d(xt)×d(xt)dt=cosxt×x=xcosxt

If y = cos1(1x1+x), then find dydx.

  1. 21+x
  2. 1x(x+1)
  3. 1xx+1
  4. 11+x

Answer (Detailed Solution Below)

Option 2 : 1x(x+1)

Trigonometric Function Question 10 Detailed Solution

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Concept:

Derivatives of Trigonometric Functions:

ddxsinx=cosx             ddxcosx=sinxddxtanx=sec2x           ddxcotx=csc2xddxsecx=tanxsecx    ddxcscx=cotxcscx

Trigonometric Formulae:

sin2x=2tanx1+tan2x

cos2x=1tan2x1+tan2x

tan2x=2tanx1tan2x

Chain Rule of Derivatives:

  • ddxf(g(x))=dd g(x)f(g(x))×ddxg(x).
  • dydx=dydu×dudx.

Calculation:

We have y = cos1(1x1+x).

Let x = tan2 z.

∴ y = cos1(1tan2z1+tan2z) = cos-1 (cos 2z) = 2z.

Now, differentiating w.r.t. z, we get:

dxdz=ddz(tan2z)=2tanzsec2z

dydz=ddz(2z)=2

Using the chain rule of derivatives, we get:

dydx=dydz×dzdx=22tanzsec2z=1tanz(1+tan2z)=1x(1+x).

Find all values of x in the interval [0, 2π] such that sin x = sin 2x?

  1. 3
  2. 5
  3. 4
  4. 2

Answer (Detailed Solution Below)

Option 2 : 5

Trigonometric Function Question 11 Detailed Solution

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Concept-

sin 2x = 2sin x cos x.

Calculation-

As sin x = sin 2x

⇒ sin 2x - sin x = 0

⇒ 2 sin x cos x - sin x = 0

⇒ sin x (2cos x - 1) = 0

So either sin x = 0 , in interval [0, 2π] when x = 0, π, 2π 

or 2cos x -1 = 0, i.e cos x = 12 in interval [0,2π] when x = π3,5π3

∴ total values of x in interval [0, 2π] is 5.

If y = (sinxcosx)sin2x, find dydx

  1. 12(secx.cotx+cosecx.tanx)
  2. 12(secx.cotxcosecx.tanx)
  3. 12(secx.tanx+cosecx.cotx)
  4. 12(secx.tanxcosecx.cotx)

Answer (Detailed Solution Below)

Option 3 : 12(secx.tanx+cosecx.cotx)

Trigonometric Function Question 12 Detailed Solution

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Concept used:

Trigonometry formula

sin 2x = 2sin x cos x

Calculation:

y = (sinxcosx)sin2x

y = (sinxcosx)2.sinx.cosx

⇒ y = sinx2.sinx.cosxcosx2.sinx.cosx

⇒ y = (12cosx)(12sinx)

⇒ y = 12 (secx - cosecx)

Differentiate both sides, we get

⇒ dydx=12[d(secx)dxd(cosecx)dx]

⇒ dydx=12(secx.tanx+cosecx.cotx)

Find dydx, if y = tan1[8x115x2]

  1. 51+25x231+9x2
  2. 51+25x2+31+9x2
  3. 81+25x2
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 51+25x2+31+9x2

Trigonometric Function Question 13 Detailed Solution

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Concept:

tan1x+tan1y=tan1[x+y1xy]

d(tan1x)dx=11+x2

Calculation:

Given: y = tan1[8x115x2]

y=tan1[5x+3x15x3x]

As we know that, tan1x+tan1y=tan1[x+y1xy]

So, y=tan1[5x+3x15x3x]= tan-1 5x + tan-1 3x

Differentiating with respect to x, we get

dydx=d(tan15x)dx+d(tan13x)dx

=51+(5x)2+31+(3x)2

=51+25x2+31+9x2

If y = sin (cos2 x2), then dydx= ?

  1. 2xcos (cos2 x2) cos x2 sin x2
  2. 4xcos (cos2 x2) cos x2 sin x2
  3. -4xcos (cos2 x2) cos x2 sin x2
  4. None of these

Answer (Detailed Solution Below)

Option 3 : -4xcos (cos2 x2) cos x2 sin x2

Trigonometric Function Question 14 Detailed Solution

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Concept:

Derivatives of Trigonometric Functions:

ddxsinx=cosx             ddxcosx=sinxddxtanx=sec2x           ddxcotx=csc2xddxsecx=tanxsecx    ddxcscx=cotxcscx

Chain Rule of Derivatives:

  • ddxf(g(x))=dd g(x)f(g(x))×ddxg(x).
  • dydx=dydu×dudx.


Calculation:

We have y = sin (cos2 x2)

Differentiating w.r.t. x, we get:

dydx = [cos (cos2 x2)] × (2 cos x2 (-sin x2)) (2x)

= -4xcos (cos2 x2) cos x2 sin x2

If f(x) = log x + 3x - 10 and g(x) = tanx then find fog'(x) 

  1. secx(cosecx3secx)
  2. secx(cosecx+3secx)
  3. secx(secx+3cosecx)
  4. secx(secx3cosecx)

Answer (Detailed Solution Below)

Option 2 : secx(cosecx+3secx)

Trigonometric Function Question 15 Detailed Solution

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Concept:

d(tanx)dx=sec2x

d(logx)dx=1x

d(x)dx=1

fog(x) = f{g(x)}

Calculation:

f(x) = log x + 3x - 10 and g(x) = tanx

fog(x) = f{g(x)} = log(tanx) + 3tanx - 10

⇒ fog'(x) = d[log(tanx)]dx+3d(tanx)dxd(10)dx

⇒ fog'(x) = 1tanxd(tanx)dx+3sec2x

⇒ fog'(x) = 1tanx×sec2x+3sec2x

⇒ fog'(x) = cosxsinx×1cos2x+3sec2x

⇒ fog'(x) = cosecx.secx+3sec2x

⇒ fog'(x) = secx(cosecx+3secx)

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