More practice questions

5. Q5. Prove that 3 cos^⁻¹ x = cos^⁻¹(4x³ − 3x) for x in [1/2, 1] (NCERT Exercise 2.2, Q2).

   Show answerHide answer

   Put x = cos θ, θ = cos^⁻¹ x. Then 4x³ − 3x = 4 cos³ θ − 3 cos θ = cos 3θ. For x in [1/2, 1], θ lies in [0, π/3], so 3θ lies in [0, π] and cos^⁻¹(cos 3θ) = 3θ = 3 cos^⁻¹ x.

6. Q6. Write in the simplest form: (i) tan^⁻¹((cos x − sin x)/(cos x + sin x)), −π/4 < x < 3π/4; (ii) tan^⁻¹(x/√(a² − x²)), |x| < a (NCERT Exercise 2.2, Q5 and Q6).

   Show answerHide answer

   (i) Since sin(π/4 − x) = (cos x − sin x)/√2 and cos(π/4 − x) = (cos x + sin x)/√2, the fraction is tan(π/4 − x) on the whole interval. (Dividing by cos x would miss x = π/2, where the fraction is −1 and tan(−π/4) = −1 as well.) π/4 − x lies in (−π/2, π/2), so the answer is π/4 − x. (ii) Put x = a sin θ with θ in (−π/2, π/2): √(a² − x²) = a cos θ, so the fraction is tan θ and the answer is θ = sin^⁻¹(x/a).

7. Q7. Write tan^⁻¹((3a²x − x³)/(a³ − 3ax²)), a > 0, −a/√3 < x < a/√3, in the simplest form (NCERT Exercise 2.2, Q7).

   Show answerHide answer

   Put x = a tan θ. The fraction becomes (3 tan θ − tan³ θ)/(1 − 3 tan² θ) = tan 3θ. The condition on x gives θ in (−π/6, π/6), so 3θ is in (−π/2, π/2). The answer is 3θ = 3 tan^⁻¹(x/a).

8. Q8. Find tan ½[sin^⁻¹(2x/(1 + x²)) + cos^⁻¹((1 − y²)/(1 + y²))], for |x| < 1, y > 0 and xy < 1 (NCERT Exercise 2.2, Q9).

   Show answerHide answer

   Put x = tan α with α in (−π/4, π/4): 2x/(1 + x²) = sin 2α and 2α lies in (−π/2, π/2), so the first term is 2α. Put y = tan β with β in (0, π/2): (1 − y²)/(1 + y²) = cos 2β with 2β in (0, π), so the second term is 2β. The expression is tan(α + β) = (x + y)/(1 − xy).

9. Q9. Find (i) cos^⁻¹(cos 13π/6); (ii) tan^⁻¹(tan 7π/6) (NCERT Miscellaneous Exercise, Q1 and Q2).

   Show answerHide answer

   (i) cos(13π/6) = cos(13π/6 − 2π) = cos(π/6), and π/6 is in [0, π]: π/6. (ii) tan(7π/6) = tan(7π/6 − π) = tan(π/6), and π/6 is in (−π/2, π/2): π/6.

10. Q10. Prove that 2 sin^⁻¹(3/5) = tan^⁻¹(24/7) (NCERT Miscellaneous Exercise, Q3).

    Show answerHide answer

    Let θ = sin^⁻¹(3/5), so tan θ = 3/4 and θ < π/4 (since 3/4 < 1). Then tan 2θ = (2 × 3/4)/(1 − 9/16) = (3/2)/(7/16) = 24/7, and 2θ lies in (0, π/2), so 2θ = tan^⁻¹(24/7).

11. Q11. Prove that (i) sin^⁻¹(8/17) + sin^⁻¹(3/5) = tan^⁻¹(77/36); (ii) cos^⁻¹(4/5) + cos^⁻¹(12/13) = cos^⁻¹(33/65) (NCERT Miscellaneous Exercise, Q4 and Q5).

    Show answerHide answer

    (i) The two angles have tangents 8/15 and 3/4, so the tangent of their sum is (8/15 + 3/4)/(1 − (8/15)(3/4)) = (77/60)/(36/60) = 77/36; both angles are acute and their sum is less than π/2 because the denominator is positive. (ii) With cos A = 4/5, sin A = 3/5 and cos B = 12/13, sin B = 5/13: cos(A + B) = (4/5)(12/13) − (3/5)(5/13) = 48/65 − 15/65 = 33/65, and A + B lies in [0, π].

12. Q12. Prove that (i) cos^⁻¹(12/13) + sin^⁻¹(3/5) = sin^⁻¹(56/65); (ii) tan^⁻¹(63/16) = sin^⁻¹(5/13) + cos^⁻¹(3/5) (NCERT Miscellaneous Exercise, Q6 and Q7).

    Show answerHide answer

    (i) With cos A = 12/13, sin A = 5/13 and sin B = 3/5, cos B = 4/5: sin(A + B) = (5/13)(4/5) + (12/13)(3/5) = 20/65 + 36/65 = 56/65, and A + B < π/2 since cos(A + B) = 48/65 − 15/65 > 0. (ii) tan(sin^⁻¹(5/13)) = 5/12 and tan(cos^⁻¹(3/5)) = 4/3; tan of the sum is (5/12 + 4/3)/(1 − 5/9) = (21/12)/(4/9) = 63/16, and the sum is acute because 5/9 < 1.

13. Q13. Prove that tan^⁻¹ √x = ½ cos^⁻¹((1 − x)/(1 + x)) for x in [0, 1] (NCERT Miscellaneous Exercise, Q8).

    Show answerHide answer

    Put x = tan² θ with θ in [0, π/4]. Then (1 − tan² θ)/(1 + tan² θ) = cos 2θ, and 2θ is in [0, π/2], so ½ cos^⁻¹(cos 2θ) = θ = tan^⁻¹ √x.

14. Q14. Solve 2 tan^⁻¹(cos x) = tan^⁻¹(2 cosec x) (NCERT Miscellaneous Exercise, Q11).

    Show answerHide answer

    cosec x must exist, so sin x ≠ 0 and |cos x| < 1. The tan^⁻¹ sum formula with x = y = t (beyond the NCERT text) gives 2 tan^⁻¹ t = tan^⁻¹(2t/(1 − t²)) for |t| < 1. With t = cos x, 2 cos x/(1 − cos² x) = 2 cos x/sin² x. Setting this equal to 2/sin x gives cos x = sin x, that is tan x = 1, so x = nπ + π/4, n ∈ Z, as in NCERT’s answers.

15. Q15. (i) sin(tan^⁻¹ x), |x| < 1, is equal to (A) x/√(1 − x²); (B) 1/√(1 − x²); (C) 1/√(1 + x²); (D) x/√(1 + x²). (ii) If sin^⁻¹(1 − x) − 2 sin^⁻¹ x = π/2, then x is (A) 0, 1/2; (B) 1, 1/2; (C) 0; (D) 1/2 (NCERT Miscellaneous Exercise, Q13 and Q14).

    Show answerHide answer

    (i) With θ = tan^⁻¹ x, cos θ > 0 and sin θ = tan θ cos θ = x/√(1 + x²): (D). (ii) Both inverse sines require x ∈ [0, 1]. For x > 0, sin^⁻¹(1 − x) < π/2 and 2 sin^⁻¹ x > 0, so the left side is strictly less than π/2. Only x = 0 works, giving sin^⁻¹ 1 − 0 = π/2. The answer is (C).

16. Q16. Prove that (i) cot^⁻¹((u + v)/(u − v)) = x/2 for x in (0, π/4), where u = √(1 + sin x) and v = √(1 − sin x); (ii) tan^⁻¹((u − v)/(u + v)) = π/4 − ½ cos^⁻¹ x for −1/√2 ≤ x ≤ 1, where now u = √(1 + x) and v = √(1 − x) (NCERT Miscellaneous Exercise, Q9 and Q10).

    Show answerHide answer

    (i) 1 ± sin x = (cos(x/2) ± sin(x/2))². On (0, π/4), cos(x/2) > sin(x/2) > 0, so the fraction is 2 cos(x/2)/(2 sin(x/2)) = cot(x/2). x/2 lies in (0, π), so the value is x/2. (ii) Put x = cos 2θ, so 2θ = cos^⁻¹ x lies in [0, 3π/4]. Then √(1 + x) = √2 cos θ and √(1 − x) = √2 sin θ, and the fraction is (cos θ − sin θ)/(cos θ + sin θ) = tan(π/4 − θ). π/4 − θ lies in [−π/8, π/4], so the value is π/4 − θ = π/4 − ½ cos^⁻¹ x.