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Class 12 Mathematics Part 1

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Previous year questions: Inverse Trigonometric Functions

Inverse Trigonometric Functions28 min readLesson 5 of 5

Q1 (CBSE Board 2023, 2 marks)

Evaluate sin inverse of sin(3pi/4) plus cos inverse of cos(pi) plus tan inverse of 1.

Answer: Reduce each term to its principal value. sin(3pi/4) is the same as sin(pi/4), and pi/4 lies in the branch of sine inverse, so the first term is pi/4. cos(pi) = -1 and cos inverse of -1 is pi. tan inverse of 1 is pi/4. Adding, pi/4 + pi + pi/4 = 3pi/2. Each reduction is recorded because the examiner gives a mark for the correct principal branch even when one arithmetic slip follows.

Q2 (CBSE Board 2024, 2 marks)

Evaluate tan inverse of 1 plus cos inverse of -1/2 plus sin inverse of -1/2.

Answer: Take the three principal values separately. tan inverse 1 = pi/4. cos inverse(-1/2) = 2pi/3 because the branch is [0, pi] and cos(2pi/3) = -1/2. sin inverse(-1/2) = -pi/6 because the branch is [-pi/2, pi/2]. The sum is pi/4 + 2pi/3 - pi/6, which is 3pi/4. Each term is checked against its own branch, so a single wrong branch flips one mark. Keep the three values in separate lines before adding them, so a sign slip in one term does not hide inside a combined fraction. Both substitutions are shown because a single wrong branch value flips the sign of the final sum.

Q3 (CBSE Board 2024, 2 marks)

Express tan inverse of cos x divided by (1 - sin x) in simplest form, for -pi/2 < x < pi/2.

Answer: Multiply numerator and denominator by (1 + sin x). The denominator becomes 1 - sin^2 x = cos^2 x. After cancelling one cos x, the expression is (1 + sin x)/cos x = sec x + tan x. The sum sec x + tan x equals tan(pi/4 + x/2). Because x lies between -pi/2 and pi/2, the angle pi/4 + x/2 lies between 0 and pi/2, inside the principal branch of tangent inverse. The simplest form is pi/4 + x/2. The sign of cos x and of 1 minus sin x is positive on the given interval, which is exactly why the half angle lands inside the branch. The half angle formula is the fastest route, and the interval check that follows is what justifies dropping the absolute value. The cancellation of one cosine factor is legal because cos x is positive on the stated interval.

Q4 (CBSE Board 2024, 2 marks)

Find the domain and range of f(x) = sin inverse of (x^2 - 4).

Answer: For sine inverse the argument must lie in [-1, 1], so -1 <= x^2 - 4 <= 1. Adding 4 gives 3 <= x^2 <= 5, hence x lies in [square root 3, square root 5] or in [minus square root 5, minus square root 3]. The range of sine inverse on its principal branch is [-pi/2, pi/2]. At x = plus or minus square root 3 the argument is minus 1, and at plus or minus square root 5 it is plus 1, so both endpoints are included. The two quadrants give two intervals for x, and both must appear in the domain, joined by the union symbol in the answer. Writing the two intervals separately first and joining them at the end avoids the common mistake of missing the negative side.

Q5 (CBSE Board 2025, 2 marks)

Find the principal value of cos inverse of -1/2 plus twice sin inverse of 1/2.

Answer: cos inverse(-1/2) has principal value 2pi/3 because cosine of 2pi/3 is -1/2. sin inverse(1/2) has principal value pi/6, so twice that is pi/3. Adding, 2pi/3 + pi/3 = pi. Keeping the two principal values separate avoids the common slip of adding the angles before reducing them to their branches. Note that the final answer is an angle value, not an inverse value, so no branch restriction applies to the sum itself. Writing the two principal values first makes the addition order irrelevant and avoids sign confusion.

Q6 (CBSE Board 2025, 2 marks)

Prove that tan inverse of square root x equals half of cos inverse of (1 - x)/(1 + x), for x in [0, 1].

Answer: Put cos inverse of (1 - x)/(1 + x) equal to theta, where theta lies in [0, pi/2] because x is between 0 and 1. Then cos theta = (1 - x)/(1 + x). Use the half angle formula tan(theta/2) = square root of ((1 - cos theta)/(1 + cos theta)). Substituting the cosine value, the numerator becomes (1 + x - 1 + x)/(1 + x) = 2x/(1 + x), and the denominator becomes (1 + x + 1 - x)/(1 + x) = 2/(1 + x). The ratio is x, so tan(theta/2) = square root x and theta/2 = tan inverse of square root x. The square root is taken as positive because the angle theta over 2 lies between 0 and pi over 4, where tangent is positive. The positivity of the square root follows from the range of the angle, so write that sentence before squaring anything. The angle theta lies in the first quadrant because x is at least zero, which is why the principal half angle is allowed without adjustment.

Q7 (CBSE Board 2026, 1 mark)

Find the principal value of sec inverse of 2 plus twice cosec inverse of -2.

Answer: sec inverse 2 equals pi/3 because sec(pi/3) = 2. cosec inverse(-2) equals -pi/6 because cosec(-pi/6) = -2. The expression becomes pi/3 + 2(-pi/6) = pi/3 - pi/3 = 0. The negative argument in the cosecant term is the only place where a sign error can enter, so reduce that term first. Both inverse values are principal values by definition, so the only work is in the sign of the cosecant term. The value zero appears because the two terms are exact opposites, so no branch adjustment is needed.

Q8 (CBSE Board 2026, 5 marks)

For p(x) = sin inverse of (1 - 2x^2), find the domain, solve p(x) = pi/6, and write the range of 2p(x) + pi/2.

Answer: Domain: -1 <= 1 - 2x^2 <= 1. Subtract 1 to get -2 <= -2x^2 <= 0, so 0 <= x^2 <= 1 and x lies in [-1, 1]. For p(x) = pi/6, the sine of pi/6 is 1/2, so 1 - 2x^2 = 1/2, giving 2x^2 = 1/2 and x^2 = 1/4, hence x = 1/2 or x = -1/2. For the range, p(x) runs over [-pi/2, pi/2], so 2p(x) runs over [-pi, pi] and adding pi/2 gives the interval [-pi/2, 3pi/2]. The domain inequality is solved before the equation, so the two solutions found later are automatically inside the domain. Both solutions satisfy the original domain, and the range step only needs the principal interval of p, not the values of x.

Q9 (CBSE Board 2026, 2 marks)

Evaluate sin inverse of tan(3pi/4).

Answer: First evaluate the inner tangent. tan(3pi/4) = -1. Then sin inverse(-1) is the principal value whose sine is -1, which is -pi/2 because the sine branch is [-pi/2, pi/2]. The order matters: evaluate the tangent first, then apply sine inverse to the resulting number. Writing the inner tangent value first makes the branch check trivial, since sine inverse of minus 1 has only one principal answer.

Q10 (CBSE SQP 2021-22, 1 mark)

For |x| < 1, the expression sin(tan inverse of x) equals which of the following? (a) x/sqrt(1 - x^2) (b) 1/sqrt(1 - x^2) (c) 1/sqrt(1 + x^2) (d) x/sqrt(1 + x^2)

Answer: Draw a right triangle with angle theta = tan inverse x. Then tan theta = x, so label the opposite side x and the adjacent side 1. The hypotenuse is sqrt(1 + x^2), and sin theta is opposite over hypotenuse, which is x/sqrt(1 + x^2). The correct option is (d). With x restricted between minus 1 and 1 the denominator is real and non-zero, so the square root expression is defined. The triangle is right angled by construction, so the hypotenuse follows from Pythagoras without extra conditions.