Exercise 2.2 — Inverse Trigonometric Functions

1. 2.2.15 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

   Prove that 3sin^{-1}x = sin^{-1}(3x-4x^3), x in [-1/2,1/2].

   Hint. Substitute x=sinθ with θ restricted so that 3θ stays inside the principal branch, then use the triple-angle sine formula.

   Let x=sinθ, so sin^{-1}x=θ, with θ restricted to [-pi/6,pi/6] since x is in [-1/2,1/2] (this keeps 3θ inside [-pi/2,pi/2]). Then 3x-4x^3=3sinθ-4sin^3θ=sin3θ, the triple-angle formula for sine. So sin^{-1}(3x-4x^3)=sin^{-1}(sin3θ)=3θ, valid since 3θ is inside the principal branch. Therefore sin^{-1}(3x-4x^3)=3θ=3sin^{-1}x.

   ✦ Identity proved: 3sin^{-1}x = sin^{-1}(3x-4x^3) for x in [-1/2,1/2].

2. 2.2.25 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

   Prove that 3cos^{-1}x = cos^{-1}(4x^3-3x), x in [1/2,1].

   Hint. Substitute x=cosθ with θ restricted so 3θ stays inside [0,pi], then use the triple-angle cosine formula.

   Let x=cosθ, so cos^{-1}x=θ, with θ restricted to [0,pi/3] since x is in [1/2,1] (this keeps 3θ inside [0,pi]). Then 4x^3-3x=4cos^3θ-3cosθ=cos3θ, the triple-angle formula for cosine. So cos^{-1}(4x^3-3x)=cos^{-1}(cos3θ)=3θ, valid since 3θ is inside the principal branch. Therefore cos^{-1}(4x^3-3x)=3θ=3cos^{-1}x.

   ✦ Identity proved: 3cos^{-1}x = cos^{-1}(4x^3-3x) for x in [1/2,1].

3. 2.2.34 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

   Write tan^{-1}[(sqrt(1+x^2)-1)/x], x != 0, in the simplest form.

   Hint. Substitute x=tanθ, so sqrt(1+x^2)=secθ, then convert the ratio to a half-angle tangent using standard identities.

   Let x=tanθ with θ in (-pi/2,pi/2), θ!=0. Then sqrt(1+x^2)=secθ (positive, since secθ>0 for θ in (-pi/2,pi/2)). The expression becomes (secθ-1)/tanθ = (1/cosθ - 1)/(sinθ/cosθ) = (1-cosθ)/sinθ = tan(θ/2), a standard half-angle identity. So tan^{-1}[(sqrt(1+x^2)-1)/x] = tan^{-1}(tan(θ/2)) = θ/2.

   ✦ (1/2)tan^{-1}x

4. 2.2.44 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

   Write tan^{-1}(sqrt((1-cos x)/(1+cos x))), 0<x<pi, in the simplest form.

   Hint. Recognise (1-cos x)/(1+cos x) as the square of a half-angle tangent.

   By the half-angle identity, (1-cos x)/(1+cos x) = tan^2(x/2). Since 0<x<pi, x/2 is in (0,pi/2), where tan(x/2)>0, so sqrt((1-cos x)/(1+cos x)) = tan(x/2). Therefore tan^{-1}(sqrt((1-cos x)/(1+cos x))) = tan^{-1}(tan(x/2)) = x/2.

   ✦ x/2

5. 2.2.54 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

   Write tan^{-1}((cos x - sin x)/(cos x + sin x)), -pi/4<x<3pi/4, in the simplest form.

   Hint. Divide numerator and denominator by cos x, then recognise the tangent-subtraction pattern with pi/4.

   Dividing numerator and denominator by cos x: (1-tan x)/(1+tan x) = (tan(pi/4)-tan x)/(1+tan(pi/4)tan x) = tan(pi/4-x), using the tangent subtraction formula. So tan^{-1}[(cos x-sin x)/(cos x+sin x)] = tan^{-1}(tan(pi/4-x)) = pi/4-x, valid because the given domain -pi/4<x<3pi/4 keeps pi/4-x inside (-pi/2,pi/2).

   ✦ pi/4 - x

6. 2.2.64 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

   Write tan^{-1}(x/sqrt(a^2-x^2)), |x|<a, in the simplest form.

   Hint. Substitute x=a sinθ, so sqrt(a^2-x^2)=a cosθ.

   Let x=a sinθ with θ in (-pi/2,pi/2). Then sqrt(a^2-x^2)=sqrt(a^2-a^2sin^2θ)=a cosθ (positive on this interval). The expression becomes tan^{-1}((a sinθ)/(a cosθ))=tan^{-1}(tanθ)=θ. Since sinθ=x/a, θ=sin^{-1}(x/a).

   ✦ sin^{-1}(x/a)

7. 2.2.75 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

   Write tan^{-1}((3a^2x-x^3)/(a^3-3ax^2)), a>0, -a/sqrt3<x<a/sqrt3, in the simplest form.

   Hint. Substitute x=a tanθ, then recognise the triple-angle tangent formula in the numerator and denominator.

   Let x=a tanθ with θ in (-pi/6,pi/6), since x/a is in (-1/sqrt3,1/sqrt3). Then 3a^2x-x^3=a^3(3tanθ-tan^3θ) and a^3-3ax^2=a^3(1-3tan^2θ), so the ratio is (3tanθ-tan^3θ)/(1-3tan^2θ)=tan3θ, the triple-angle tangent formula. So the expression becomes tan^{-1}(tan3θ)=3θ, valid since 3θ is inside (-pi/2,pi/2). Since tanθ=x/a, θ=tan^{-1}(x/a).

   ✦ 3tan^{-1}(x/a)

8. 2.2.83 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

   Find the value of tan^{-1}[2cos(2sin^{-1}(1/2))].

   Hint. Work from the innermost expression outward: evaluate sin^{-1}(1/2) first.

   sin^{-1}(1/2)=pi/6, so 2sin^{-1}(1/2)=pi/3. cos(pi/3)=1/2. So 2cos(pi/3)=2(1/2)=1. Finally tan^{-1}(1)=pi/4.

   ✦ pi/4

9. 2.2.96 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

   Find the value of tan{(1/2)[sin^{-1}(2x/(1+x^2)) + cos^{-1}((1-y^2)/(1+y^2))]}, |x|<1, y>0, xy<1.

   Hint. Substitute x=tanα and y=tanβ, then recognise both inner expressions as double-angle formulas.

   Let x=tanα with α in (-pi/4,pi/4) (since |x|<1), and y=tanβ with β in (0,pi/2) (since y>0). Then 2x/(1+x^2)=sin2α (valid since 2α is in (-pi/2,pi/2)), so sin^{-1}(2x/(1+x^2))=2α. Also (1-y^2)/(1+y^2)=cos2β (valid since 2β is in (0,pi)), so cos^{-1}((1-y^2)/(1+y^2))=2β. The expression becomes tan[(1/2)(2α+2β)]=tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=(x+y)/(1-xy).

   ✦ (x+y)/(1-xy)

10. 2.2.102 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

    Find the value of sin^{-1}(sin(2pi/3)).

    Hint. 2pi/3 is outside the principal branch — find the equivalent angle inside it using the supplementary-angle relation for sine.

    2pi/3 is not in [-pi/2,pi/2]. Since sin(2pi/3)=sin(pi-2pi/3)=sin(pi/3), and pi/3 is inside [-pi/2,pi/2], sin^{-1}(sin(2pi/3))=sin^{-1}(sin(pi/3))=pi/3.

    ✦ pi/3

11. 2.2.112 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

    Find the value of tan^{-1}(tan(3pi/4)).

    Hint. 3pi/4 is outside the principal branch — use the period-pi property of tangent to find the equivalent angle inside it.

    3pi/4 is not in (-pi/2,pi/2). Since tangent has period pi, tan(3pi/4)=tan(3pi/4-pi)=tan(-pi/4), and -pi/4 is inside (-pi/2,pi/2). So tan^{-1}(tan(3pi/4))=-pi/4.

    ✦ -pi/4

12. 2.2.124 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

    Find the value of tan(sin^{-1}(3/5) + cot^{-1}(3/2)).

    Hint. Extract the tangent ratio from each inverse trig term using a right triangle, then apply the tangent addition formula.

    Let alpha=sin^{-1}(3/5): sin(alpha)=3/5, cos(alpha)=4/5, tan(alpha)=3/4 (alpha in the first quadrant). Let beta=cot^{-1}(3/2): cot(beta)=3/2, tan(beta)=2/3 (beta in the first quadrant, since cot(beta)>0). tan(alpha+beta) = (tan alpha+tan beta)/(1-tan alpha tan beta) = (3/4+2/3)/(1-(3/4)(2/3)) = (17/12)/(1/2) = 17/6.

    ✦ 17/6

13. 2.2.132 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

    cos^{-1}(cos(7pi/6)) is equal to (A) 7pi/6 (B) 5pi/6 (C) pi/3 (D) pi/6.

    Hint. 7pi/6 is outside [0,pi] — use that cosine is even and periodic to find the equivalent angle inside the branch.

    7pi/6 is not in [0,pi]. cos(7pi/6)=cos(pi+pi/6)=-cos(pi/6)=-sqrt3/2. Also cos(5pi/6)=cos(pi-pi/6)=-cos(pi/6)=-sqrt3/2, matching the same value, and 5pi/6 is inside [0,pi]. So cos^{-1}(cos(7pi/6))=5pi/6.

    ✦ (B) 5pi/6

14. 2.2.143 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

    sin(pi/3 - sin^{-1}(-1/2)) is equal to (A) 1/2 (B) 1/3 (C) 1/4 (D) 1.

    Hint. Evaluate the inner inverse sine first, then simplify the angle before taking sine.

    sin^{-1}(-1/2)=-pi/6. So pi/3-(-pi/6)=pi/3+pi/6=2pi/6+pi/6=3pi/6=pi/2. sin(pi/2)=1.

    ✦ (D) 1

15. 2.2.153 marksNCERT Class 12 Mathematics, Inverse Trigonometric Functions, Reprint 2026-27

    tan^{-1}(sqrt3) - cot^{-1}(-sqrt3) is equal to (A) pi (B) -pi/2 (C) 0 (D) 2sqrt3.

    Hint. Evaluate each principal value, remembering the sign convention for cot^{-1} of a negative number.

    tan^{-1}(sqrt3)=pi/3. For cot^{-1}(-sqrt3): need cot(theta)=-sqrt3 with theta in (0,pi), giving theta=5pi/6 (since cot(5pi/6)=cos(5pi/6)/sin(5pi/6)=(-sqrt3/2)/(1/2)=-sqrt3). So the expression is pi/3-5pi/6 = 2pi/6-5pi/6 = -3pi/6 = -pi/2.

    ✦ (B) -pi/2

Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh102.pdf) — Exercise 2.1 (14 questions), Exercise 2.2 (15 questions), plus the chapter's Miscellaneous Exercise (14 questions), 43 questions total. Exercise pages were rendered as 300dpi images to read the stacked-fraction inverse-trig notation accurately, since raw text extraction badly garbled it. The old stub had no solutions file at all and collapsed the three real exercises into one invented 21-question group; it also taught a numbered list of 'properties' (like sin^{-1}x+cos^{-1}x=pi/2) that the current edition does not box anywhere — confirmed by reading the full text of Section 2.3, which goes directly from a one-line recap into three worked examples that all use the same sinθ/cosθ/tanθ/secθ substitution technique rather than stating identities to memorise. Every proof-based answer in this file was independently re-derived from the substitution technique (not copied from a key), and Miscellaneous Q14 was specifically checked for the extraneous root x=1/2 that the squaring-equivalent step introduces but that fails the original equation.. Questions are referenced from the NCERT textbook for identification.