68 free MCQs on Inverse Trigonometric Functions with worked answers and explanations. Restricting trig functions to make them invertible, the principal value branches of sin-inverse, cos-inverse, tan-inverse, and friends, and the key identities relating them.
Take the timed Inverse Trigonometric Functions chapterwise test →Below are 68 practice questions on Inverse Trigonometric Functions, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Inverse Trigonometric Functions notes.
The graph of sin⁻¹x is confined to a narrow domain [-1,1] (since sine itself only takes values in that range) and range [-π/2,π/2] (the principal value branch chosen to make sine one-one and invertible there).
Q1.
The principal value range of sin-1 x is:
Answer: B. [-pi/2, pi/2]
Why: sin-1 x has principal value branch [-pi/2, pi/2], with domain [-1,1].
Q2.
The principal value range of cos-1 x is:
Answer: B. [0, pi]
Why: cos-1 x has principal value branch [0, pi], with domain [-1,1].
Q3.
The principal value range of tan-1 x is:
Answer: B. (-pi/2, pi/2)
Why: tan-1 x has principal value branch (-pi/2, pi/2), open interval since tan is undefined at the endpoints, with domain all real numbers.
Q4.
Why must the domain of sin x be restricted before defining sin-1 x?
Answer: B. sin x is periodic and not one-one over all reals, so it has no inverse without restriction
Why: Since trig functions repeat periodically, they are not one-one on their full domain; restricting to a principal branch makes them invertible.
Q5.
sin-1(1/2) =
Answer: A. pi/6
Why: sin(pi/6) = 1/2, and pi/6 lies within the principal range, so sin-1(1/2) = pi/6.
Q6.
cos-1(1/2) =
Answer: C. pi/3
Why: cos(pi/3) = 1/2, and pi/3 lies within [0,pi], so cos-1(1/2) = pi/3.
Q7.
sin-1 x + cos-1 x =
Answer: B. pi/2
Why: This is a fundamental identity: sin-1 x + cos-1 x = pi/2 for all x in [-1,1].
Q8.
tan-1 x + cot-1 x =
Answer: B. pi/2
Why: This identity holds for all real x: tan-1 x + cot-1 x = pi/2.
Q9.
sin-1(-x) equals:
Answer: B. -sin-1 x
Why: sin-1 is an odd function: sin-1(-x) = -sin-1 x.
Q10.
cos-1(-x) equals:
Answer: C. pi - cos-1 x
Why: cos-1 is not odd; the correct identity is cos-1(-x) = pi - cos-1 x.
Q11.
The domain of sec-1 x is:
Answer: C. R minus the open interval (-1,1)
Why: sec-1 x is defined for |x| >= 1, i.e., the domain excludes the open interval (-1,1).
Q12.
tan-1(sqrt(3)) =
Answer: C. pi/3
Why: tan(pi/3) = sqrt(3), and pi/3 lies in the principal range, so tan-1(sqrt3) = pi/3.
Q13.
sec-1 x + cosec-1 x =
Answer: B. pi/2
Why: For |x| >= 1, sec-1 x + cosec-1 x = pi/2, analogous to the sin-1/cos-1 identity.
Q14.
The range of cot-1 x excludes which value?
Answer: D. It excludes both 0 and pi as open endpoints
Why: The principal value range of cot-1 x is the open interval (0, pi), so both endpoints 0 and pi are excluded.
Q15.
The value of sin⁻¹(0) is:
Answer: A. 0
Why: sin 0 = 0, so sin⁻¹(0) = 0.
Q16.
The value of cos⁻¹(1) is:
Answer: A. 0
Why: cos 0 = 1, so cos⁻¹(1) = 0.
Q17.
The value of tan⁻¹(0) is:
Answer: A. 0
Why: tan 0 = 0, so tan⁻¹(0) = 0.
Q18.
The function sin⁻¹(x) is also written as:
Answer: A. arcsin x
Why: sin⁻¹(x) and arcsin x denote the same inverse-sine function.
Q19.
The value of sin⁻¹(1) is:
Answer: A. exactly π/2
Why: sin(π/2) = 1, so sin⁻¹(1) = π/2.
Q20.
The domain of the function sin⁻¹(x) is:
Answer: A. [−1, 1]
Why: Since sine only outputs values in [−1, 1], its inverse accepts inputs in [−1, 1].
Q21.
Evaluate sin-1(sin(3pi/4)).
Answer: B. pi/4
Why: sin(3pi/4) = sin(pi/4) = 1/sqrt(2). Since 3pi/4 is outside [-pi/2,pi/2], the principal value is sin-1(1/sqrt2) = pi/4.
Q22.
Evaluate cos-1(cos(4pi/3)).
Answer: B. 2pi/3
Why: cos(4pi/3) = -1/2. Since 4pi/3 is outside [0,pi], we find the principal value: cos-1(-1/2) = 2pi/3.
Q23.
Simplify tan-1(1/2) + tan-1(1/3).
Answer: A. pi/4
Why: Using tan-1 x + tan-1 y = tan-1[(x+y)/(1-xy)] with x=1/2, y=1/3: (1/2+1/3)/(1-1/6) = (5/6)/(5/6) = 1. So the sum is tan-1(1) = pi/4 (valid since xy=1/6<1).
Q24.
Find the value of sin(cos-1(3/5)).
Answer: B. 4/5
Why: If theta = cos-1(3/5), then cos(theta) = 3/5, so the opposite side is sqrt(25-9) = 4, giving sin(theta) = 4/5.
Q25.
Evaluate tan(sin-1(3/5)).
Answer: A. 3/4
Why: If theta = sin-1(3/5), then sin(theta)=3/5 and cos(theta) = 4/5 (right triangle with opposite 3, hyp 5, adjacent 4). tan(theta) = 3/4.
Q26.
Simplify 2 tan-1(1/2) using the double angle identity 2tan-1 x = sin-1(2x/(1+x<sup>2</sup>)).
Answer: A. sin-1(4/5)
Why: With x=1/2: 2x/(1+x<sup>2</sup>) = 1/(1+1/4) = 1/(5/4) = 4/5. So 2tan-1(1/2) = sin-1(4/5) (valid since |x|<=1).
Q27.
Express cot-1(-x) in terms of cot-1 x.
Answer: C. pi - cot-1 x
Why: Like cos-1, cot-1 is not odd; the identity is cot-1(-x) = pi - cot-1 x.
Q28.
If sin-1 x = pi/3, find the value of cos-1 x.
Answer: A. pi/6
Why: Using sin-1 x + cos-1 x = pi/2: cos-1 x = pi/2 - pi/3 = pi/6.
Q29.
Solve for x: tan-1(2x) + tan-1(3x) = pi/4.
Answer: A. x = 1/6
Why: tan-1(2x)+tan-1(3x) = tan-1[(5x)/(1-6x<sup>2</sup>)] = pi/4 implies 5x/(1-6x<sup>2</sup>)=1, so 5x = 1-6x<sup>2</sup>, giving 6x<sup>2</sup>+5x-1=0, (6x-1)(x+1)=0, so x=1/6 (x=-1 is rejected as it gives a negative argument issue with the formula's validity).
Q30.
Express tan-1(cos x / (1 + sin x)) in simplified principal value form for x in (-pi/2, pi/2).
Answer: A. pi/4 - x/2
Why: Using half-angle substitutions, cos x/(1+sin x) simplifies to tan(pi/4 - x/2), so the expression equals pi/4 - x/2 within the valid principal range.
Q31.
Find the principal value of cosec-1(-1).
Answer: B. -pi/2
Why: cosec(-pi/2) = -1, and -pi/2 lies in the principal range [-pi/2,pi/2] excluding 0, so cosec-1(-1) = -pi/2.
Q32.
Find the principal value of cot-1(-1).
Answer: B. 3pi/4
Why: cot(3pi/4) = -1, and 3pi/4 lies in the principal range (0,pi), so cot-1(-1) = 3pi/4.
Q33.
Evaluate tan-1(tan(2pi/3)).
Answer: B. -pi/3
Why: tan(2pi/3) = -sqrt(3). Since 2pi/3 is outside (-pi/2,pi/2), find principal value: tan-1(-sqrt3) = -pi/3.
Q34.
Simplify 2 tan-1(1/3) using the double angle identity 2tan-1 x = tan-1(2x/(1-x<sup>2</sup>)).
Answer: A. tan-1(3/4)
Why: With x=1/3: 2x/(1-x<sup>2</sup>) = (2/3)/(1-1/9) = (2/3)/(8/9) = (2/3)(9/8) = 3/4. So 2tan-1(1/3) = tan-1(3/4) (valid since |x|<1).
Q35.
The principal value range of sin⁻¹(x) is:
Answer: A. [−π/2, π/2]
Why: The principal branch of arcsine is [−π/2, π/2].
Q36.
The principal value range of cos⁻¹(x) is:
Answer: A. [0, π]
Why: The principal branch of arccosine is [0, π].
Q37.
For all x in [−1, 1], sin⁻¹(x) + cos⁻¹(x) equals:
Answer: A. π/2
Why: sin⁻¹(x) + cos⁻¹(x) = π/2 is a standard identity.
Q38.
For all real x, tan⁻¹(x) + cot⁻¹(x) equals:
Answer: A. π/2
Why: tan⁻¹(x) + cot⁻¹(x) = π/2.
Q39.
The value of cos⁻¹(1/2) is:
Answer: A. π/3
Why: cos(π/3) = 1/2, so cos⁻¹(1/2) = π/3.
Q40.
The value of tan⁻¹(1) is:
Answer: A. π/4
Why: tan(π/4) = 1, so tan⁻¹(1) = π/4.
Q41.
Solve: sin-1(x) + sin-1(2x) = pi/3. Approximate the smaller positive root region check, given x must satisfy domain constraints.
Answer: B. x = sqrt(3)/(2sqrt(7)) is the value satisfying domain and equation
Why: Setting sin-1(2x) = pi/3 - sin-1(x) and taking sine of both sides with the addition formula leads, after squaring and simplifying (7x<sup>2</sup> = 3/4 form), to x = sqrt(3)/(2sqrt(7)), which satisfies both the equation and domain |2x|<=1.
Q42.
Simplify tan-1[(sqrt(1+x<sup>2</sup>) - 1)/x] for x > 0 in terms of tan-1 x.
Answer: A. (1/2) tan-1 x
Why: Substituting x = tan(theta) with theta in (0,pi/2), (sqrt(1+x<sup>2</sup>)-1)/x = (sec(theta)-1)/tan(theta) = tan(theta/2), so the expression equals theta/2 = (1/2)tan-1 x.
Q43.
If cos-1 x + cos-1 y + cos-1 z = 3pi (the maximum possible sum), what must be true of x, y, z?
Answer: C. x = y = z = -1
Why: Since each cos-1 term has a maximum value of pi (attained only when the argument is -1), the sum equals 3pi only when x=y=z=-1.
Q44.
Evaluate tan-1(1) + tan-1(2) + tan-1(3).
Answer: A. pi
Why: tan-1(2)+tan-1(3) = pi + tan-1[(2+3)/(1-6)] = pi + tan-1(-1) = pi - pi/4 = 3pi/4 (since xy=6>1 and both positive, add pi). Then tan-1(1) + 3pi/4 = pi/4 + 3pi/4 = pi.
Q45.
Find the value of cos[2cos-1(3/5) ] using the double angle formula cos(2theta) = 2cos<sup>2</sup>(theta) - 1.
Answer: A. -7/25
Why: Let θ=cos⁻¹(3/5), so cosθ=3/5 (θ∈[0,π], principal range). Apply cos2θ=2cos²θ−1=2(9/25)−1=18/25−25/25=−7/25. Since 2θ∈[0,2π], the result −7/25 is valid. Answer: −7/25.
Q46.
Solve for x: 2 tan-1(cos x) = tan-1(2 cosec x), for x in (0, pi/2).
Answer: A. x = pi/4
Why: At x = pi/4: cos(pi/4) = 1/sqrt2, cosec(pi/4) = sqrt2. LHS: 2tan-1(1/sqrt2). RHS: tan-1(2sqrt2). Using the double angle formula 2tan-1(1/sqrt2) = tan-1[2(1/sqrt2)/(1-1/2)] = tan-1[sqrt2/(1/2)] = tan-1(2sqrt2), matching RHS, confirming x=pi/4 works.
Q47.
If tan-1 x + tan-1 y + tan-1 z = pi and x, y, z > 0, which relation among x, y, z holds?
Answer: A. x + y + z = xyz
Why: Let A=tan⁻¹x, B=tan⁻¹y, C=tan⁻¹z with A+B+C=π. Then A+B=π−C, so tan(A+B)=tan(π−C)=−tanC. Expanding: (x+y)/(1−xy)=−z → x+y=−z+xyz → x+y+z=xyz. Answer: x+y+z=xyz.
Q48.
Evaluate sin[cos-1(4/5) + tan-1(2/3)] using compound angle expansion.
Answer: C. 17/(5sqrt13)
Why: For cos-1(4/5): cos=4/5, sin=3/5. For tan-1(2/3): in a right triangle with opposite 2, adjacent 3, hyp sqrt13, so sin=2/sqrt13, cos=3/sqrt13. sin(A+B)=sinAcosB+cosAsinB = (3/5)(3/sqrt13)+(4/5)(2/sqrt13) = (9+8)/(5sqrt13) = 17/(5sqrt13).
Q49.
If sin<sup>-1</sup>(x) - cos<sup>-1</sup>(x) = pi/6, find the value of x.
Answer: D. sqrt(3)/2
Why: Use complementary identity: sin⁻¹x+cos⁻¹x=π/2. Add both equations: 2sin⁻¹x=π/2+π/6=2π/3 → sin⁻¹x=π/3 → x=sin(π/3)=√3/2. Check: cos⁻¹(√3/2)=π/6, difference=π/3−π/6=π/6 ✓.
Q50.
Find the value of tan[2 tan<sup>-1</sup>(1/5) - pi/4].
Answer: A. -7/17
Why: Let theta = tan<sup>-1</sup>(1/5); tan 2theta = 5/12 by the double-angle formula. Then tan(2theta - pi/4) = (tan2theta - 1)/(1 + tan2theta) = -7/17.
Q51.
The value of sin⁻¹(−1/2) is:
Answer: A. −π/6
Why: Since arcsine is odd and sin⁻¹(1/2) = π/6, we get sin⁻¹(−1/2) = −π/6.
Q52.
The value of cos⁻¹(−1/2) is:
Answer: A. 2π/3
Why: cos⁻¹(−x) = π − cos⁻¹(x) = π − π/3 = 2π/3.
Q53.
For x in [−1, 1], sin(sin⁻¹ x) equals:
Answer: A. x
Why: Applying sine to arcsine returns the original value x.
Q54.
The value of tan⁻¹(√3) is:
Answer: A. π/3
Why: tan(π/3) = √3, so tan⁻¹(√3) = π/3.
Q55.
If sin⁻¹(x) = θ, then x equals:
Answer: A. sin θ
Why: By definition of the inverse function, x = sin θ.
Q56.
The expression 2 tan⁻¹(x) can be written as tan⁻¹ of:
Answer: A. 2x/(1 − x²)
Why: 2 tan⁻¹(x) = tan⁻¹(2x/(1 − x²)) for |x| < 1.
Q57.
The value of cos(cos⁻¹(0.5)) is:
Answer: A. 0.5
Why: cos and cos⁻¹ are inverses, so cos(cos⁻¹(0.5)) = 0.5.
Q58.
For suitable x, sec⁻¹(x) + cosec⁻¹(x) equals:
Answer: A. π/2
Why: sec⁻¹(x) + cosec⁻¹(x) = π/2, analogous to the other complementary identities.
Q59.
The value of sin⁻¹(sin(π/3)) is:
Answer: A. π/3
Why: Since π/3 lies in the principal range [−π/2, π/2], sin⁻¹(sin(π/3)) = π/3.
Q60.
The value of tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) is:
Answer: A. π
Why: This is a well-known identity: tan⁻¹1 + tan⁻¹2 + tan⁻¹3 = π.
Q61.
The value of arctan1 + arctan2 + arctan3 is:
Answer: B. π
Why: This is a classic identity: the three inverse tangents sum to π.
Q62.
The value of arcsin(sin(2π/3)) is:
Answer: B. π/3
Why: Since 2π/3 lies outside [−π/2, π/2], arcsin(sin(2π/3)) = π − 2π/3 = π/3.
Q63.
The value of arccos(cos(7π/6)) is:
Answer: B. 5π/6
Why: 7π/6 lies outside [0, π]; the principal value is 2π − 7π/6 = 5π/6.
Q64.
The value of arctan(1/2) + arctan(1/3) is:
Answer: B. π/4
Why: Using the sum formula, ((1/2 + 1/3)/(1 − 1/6)) = 1, so the sum is arctan1 = π/4.
Q65.
The principal-value range of arcsec x is:
Answer: B. [0, π] excluding π/2
Why: arcsec takes values in [0, π] but never π/2.
Q66.
The value of sin(2·arctan(3/4)) is:
Answer: A. 24/25
Why: With tanθ = 3/4, sin2θ = 2tanθ/(1 + tan²θ) = (3/2)/(25/16) = 24/25.
Q67.
For x > 0, arctan x + arctan(1/x) equals:
Answer: C. π/2
Why: For positive x these complementary angles sum to π/2.
Q68.
The principal value of arccot(−1) is:
Answer: B. 3π/4
Why: arccot has range (0, π); arccot(−1) = 3π/4.