Inverse Trigonometric Functions — Questions & Answers

Sample questions from Inverse Trigonometric Functions

1. Q1. The value of cos(tan⁻¹(3/4) − sin⁻¹(5/13)) is:

   • A.56/65
   • B.33/65
   • C.16/65
   • D.63/65✓
   Solution

   A = tan⁻¹(3/4): sin A = 3/5, cos A = 4/5. B = sin⁻¹(5/13): sin B = 5/13, cos B = 12/13. cos(A−B) = (4/5)(12/13) + (3/5)(5/13) = 48/65 + 15/65 = 63/65.

2. Q2. The domain of f(x) = cos⁻¹(3x + 1) is:

   • A.[−1/3, 2/3]
   • B.[−2/3, 0]✓
   • C.[0, 2/3]
   • D.[−2/3, 2/3]
   Solution

   We need −1 ≤ 3x + 1 ≤ 1, so −2 ≤ 3x ≤ 0, giving −2/3 ≤ x ≤ 0.

3. Q3. The number of real solutions of sin⁻¹(x) + cos⁻¹(x) = tan⁻¹(x) is:

   • A.0
   • B.1✓
   • C.2
   • D.Infinite
   Solution

   B is correct. sin⁻¹(x)+cos⁻¹(x)=π/2 for all x∈[−1,1]. So the equation becomes π/2 = tan⁻¹(x), which requires x=tan(π/2), which is undefined (tan diverges at π/2). However, for x∈R, tan⁻¹(x)<π/2, so tan⁻¹(x)=π/2 has no solution. So there are 0 solutions... but wait — tan⁻¹(x)→π/2 as x→∞ but never equals π/2. And for x∉[−1,1], sin⁻¹(x) is undefined. So the equation has 0 real solutions. Checking again: if x∈[−1,1], lhs=π/2, and tan⁻¹(x)≤tan⁻¹(1)=π/4<π/2. So 0 solutions.

4. Q4. The simplified form of tan⁻¹(sqrt((1 − cos x)/(1 + cos x))) for 0 < x < π is:

   • A.x
   • B.x/2✓
   • C.π/2 − x/2
   • D.2x
   Solution

   Using 1 − cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2), the radicand is tan²(x/2). For 0 < x < π, tan(x/2) > 0, so the expression equals tan⁻¹(tan(x/2)) = x/2.

5. Q5. For 0 < x < π/2, the expression tan⁻¹(cos x/(1 − sin x)) equals:

   • A.π/4 − x/2
   • B.x/2
   • C.π/2 − x
   • D.π/4 + x/2✓
   Solution

   Write cos x = cos²(x/2) − sin²(x/2) and 1 − sin x = (cos(x/2) − sin(x/2))². Cancelling one factor gives (cos(x/2) + sin(x/2))/(cos(x/2) − sin(x/2)) = tan(π/4 + x/2). Since π/4 + x/2 ∈ (π/4, π/2) lies in the principal range, the value is π/4 + x/2.

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