📐 Mathematics · Class 12 · JEE
Inverse Trigonometric Functions - Practice Questions with Answers 68 free MCQs on Inverse Trigonometric Functions, each with its own worked answer and explanation. 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 → 68 practice questions on Inverse Trigonometric Functions , sorted Easy → Hard. Try each one first, then open its answer page for the worked explanation. Want the full theory first? Read the Inverse Trigonometric Functions notes .
y = sin⁻¹x: Principal Value Branch x y x=-1 x=1 y=-π/2 y=π/2 Domain restricted to [-1,1]; range restricted to [-π/2,π/2] - this restriction is what makes the inverse exist 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).
Easy - 20 questions Q4.
Why must the domain of sin x be restricted before defining sin-1 x?
A sin x fails to be continuous at certain rational multiples of pi, breaking the inverse constructionB sin x is periodic and not one-one over all reals, so it has no inverse without restrictionC sin x has no defined range over the real numbers, so an inverse formula cannot be writtenD sin x is always positive for every real input value, leaving no negative outputs to invertShow answer & explanation →
Q11.
The domain of sec-1 x is:
A All real numbers without any restrictionB The closed interval from -1 to 1C R minus the open interval (-1,1)D Positive real numbers greater than zeroShow answer & explanation →
Q14.
The range of cot-1 x excludes which value?
A 0, since the range starts strictly above zeroB pi/2, a value that cot-1 x does not actually reachC pi, a value that cot-1 x does not actually attainD It excludes both 0 and pi as open endpointsShow answer & explanation →
Medium - 20 questions Q26.
Simplify 2 tan-1(1/2) using the double angle identity 2tan-1 x = sin-1(2x/(1+x<sup>2</sup>)).
A sin-1(4/5)B sin-1(3/5)C sin-1(1)D sin-1(1/2)Show answer & explanation →
Q30.
Express tan-1(cos x / (1 + sin x)) in simplified principal value form for x in (-pi/2, pi/2).
A pi/4 - x/2B pi/4 + x/2C x/2D pi/2 - xShow answer & explanation →
Q34.
Simplify 2 tan-1(1/3) using the double angle identity 2tan-1 x = tan-1(2x/(1-x<sup>2</sup>)).
A tan-1(3/4)B tan-1(2/3)C tan-1(1/2)D tan-1(4/3)Show answer & explanation →
Hard - 28 questions Q41.
Solve: sin-1(x) + sin-1(2x) = pi/3. Approximate the smaller positive root region check, given x must satisfy domain constraints.
A x = 1/(2sqrt(7)), obtained by dropping the sqrt(3) factor in the derivationB x = sqrt(3)/(2sqrt(7)) is the value satisfying domain and equationC x = 1/2, which fails the domain constraint required for sin-1(2x)D x = 1/sqrt(7), obtained from an algebra slip in clearing the radicalShow answer & explanation →
Q42.
Simplify tan-1[(sqrt(1+x<sup>2</sup>) - 1)/x] for x > 0 in terms of tan-1 x.
A (1/2) tan-1 xB 2 tan-1 xC tan-1 xD (1/2) tan-1(1/x)Show answer & explanation →
Q43.
If cos-1 x + cos-1 y + cos-1 z = 3pi (the maximum possible sum), what must be true of x, y, z?
A x = y = z = 0B x = y = z = 1C x = y = z = -1D x + y + z = 0Show answer & explanation →
Q45.
Find the value of cos[2cos-1(3/5) ] using the double angle formula cos(2theta) = 2cos<sup>2</sup>(theta) - 1.
A -7/25B 7/25C 18/25D -18/25Show answer & explanation →
Q46.
Solve for x: 2 tan-1(cos x) = tan-1(2 cosec x), for x in (0, pi/2).
A x = pi/4B x = pi/6C x = pi/3D x = pi/2Show answer & explanation →
Q47.
If tan-1 x + tan-1 y + tan-1 z = pi and x, y, z > 0, which relation among x, y, z holds?
A x + y + z = xyzB xyz = 1C x + y + z = 0D xy + yz + zx = 1Show answer & explanation →
Q48.
Evaluate sin[cos-1(4/5) + tan-1(2/3)] using compound angle expansion.
A (8+3sqrt(13))/(5sqrt(13))B (8-3sqrt(13))/(5sqrt(13))C 17/(5sqrt13)D 6/(5sqrt13)Show answer & explanation →