Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
How to Use This Page
This page is your complete study companion for Chapter 4 — Inverse Trigonometric Functions. Start with the Key Concepts table to refresh definitions and ranges, then work through the MCQs, fill-in-the-blanks, and true-or-false sections for quick practice. Use the match-the-following exercise to strengthen associations. The brief and detailed answer sections mirror the 2-mark and 5-mark questions you will face in the board exam. Finally, review the Common Mistakes and Exam Checklist before every test to avoid losing marks.
Chapter 4 — Exercise Solutions Guide
Inverse trigonometry is mostly about using the correct identity and respecting the principal value range.
Key Concepts (Quick Recap)
| Term | Definition | ||
|---|---|---|---|
| Inverse trigonometric function | The inverse of a trigonometric function restricted to its principal value domain | ||
| sin⁻¹ x | Gives the angle whose sine is x; domain [-1, 1], range [-pi/2, pi/2] | ||
| cos⁻¹ x | Gives the angle whose cosine is x; domain [-1, 1], range [0, pi] | ||
| tan⁻¹ x | Gives the angle whose tangent is x; domain all real numbers, range (-pi/2, pi/2) | ||
| cot⁻¹ x | Gives the angle whose cotangent is x; domain all real numbers, range (0, pi) | ||
| sec⁻¹ x | Gives the angle whose secant is x; domain (-inf, -1] union [1, inf), range [0, pi] minus {pi/2} | ||
| cosec⁻¹ x | Gives the angle whose cosecant is x; domain (-inf, -1] union [1, inf), range [-pi/2, pi/2] minus {0} | ||
| Principal value | The unique value of the inverse function in its specified range | ||
| Complementary identity | sin⁻¹ x + cos⁻¹ x = pi/2 for all x in [-1, 1] | ||
| Addition formula | tan⁻¹ x + tan⁻¹ y = tan⁻¹((x + y)/(1 - xy)) when xy < 1 | ||
| Subtraction formula | tan⁻¹ x - tan⁻¹ y = tan⁻¹((x - y)/(1 + xy)) | ||
| Double angle | 2 tan⁻¹ x = tan⁻¹(2x/(1 - x²)) when \ | x\ | < 1 |
| sin⁻¹(sin theta) | Equals theta only when theta is in [-pi/2, pi/2]; otherwise adjust |
Principal Value Ranges
| Function | Principal range |
|---|---|
| sin⁻¹ x | [-pi/2, pi/2] |
| cos⁻¹ x | [0, pi] |
| tan⁻¹ x | (-pi/2, pi/2) |
| cot⁻¹ x | (0, pi) |
| sec⁻¹ x | [0, pi] - {pi/2} |
| cosec⁻¹ x | [-pi/2, pi/2] - {0} |
Important Identities
I. Choose the Best Answer (MCQs)
II. Fill in the Blanks
| 9. 2 tan⁻¹ x = sin⁻¹(______) for | x | <= 1. |
|---|
III. True or False
1. sin⁻¹ x always gives a value in the range [0, pi]. False. Correction: sin⁻¹ x gives a value in [-pi/2, pi/2]. It is cos⁻¹ x that gives a value in [0, pi].
2. tan⁻¹ x and cot⁻¹ x have the same domain (all real numbers). True. (Both functions are defined for all real x, though their ranges differ.)
3. cos⁻¹(cos theta) = theta for every real theta. False. Correction: cos⁻¹(cos theta) = theta only when theta is in [0, pi]. Outside this range, the value must be adjusted.
4. sin⁻¹(-x) = -sin⁻¹(x) for all x in [-1, 1]. True. (sin⁻¹ is an odd function: sin⁻¹(-x) = -sin⁻¹ x.)
5. tan⁻¹ x + cot⁻¹ x = pi/2 for all real x. True. (This identity holds for every real value of x, not just positive values.)
6. The value of sin⁻¹(2) is defined. False. Correction: sin⁻¹(2) is not defined because 2 is outside the domain [-1, 1] of sin⁻¹.
7. sec⁻¹ x = cos⁻¹(1/x) for all valid x. True. (This relation holds for x >= 1 or x <= -1, which are the valid domains of sec⁻¹.)
True. (This is the double angle formula for sin⁻¹, valid when the argument lies in the principal range.)
| 8. 2 sin⁻¹ x = sin⁻¹(2x sqrt(1-x²)) for | x | <= 1/sqrt(2). |
|---|
IV. Match the Following
| Column A | Column B |
|---|---|
| 1. sin⁻¹(1) | (c) pi/2 |
| 2. cos⁻¹(0) | (f) pi/2 |
| 3. tan⁻¹(sqrt(3)) | (a) pi/3 |
| 4. sin⁻¹ x + cos⁻¹ x | (e) pi/2 |
| 5. tan⁻¹ x + cot⁻¹ x | (e) pi/2 |
| 6. cos⁻¹(-1) | (b) pi |
| 7. sin⁻¹(sin 5pi/6) | (g) pi/6 |
| 8. tan⁻¹(1) | (d) pi/4 |
Standard Solution Pattern
1. Identify the inverse function. 2. Check whether an identity applies. 3. Convert expression to one inverse function if possible. 4. Apply range condition. 5. Write final value in radians.
Worked Pattern
Simplify:
Using identity:
This is valid for -1 <= x <= 1.
Worked Examples
Example 1 — Principal values
Example 2 — sin⁻¹(sin theta) outside the principal range
Evaluate sin⁻¹(sin 2pi/3).
Example 3 — Sum identity
Show tan⁻¹1 + tan⁻¹2 + tan⁻¹3 = pi.
Example 4 — Using complementary identity
Evaluate sin⁻¹(3/5) + cos⁻¹(3/5).
Example 5 — Simplification using addition formula
Find the value of tan⁻¹(1/2) + tan⁻¹(1/3).
V. Answer Briefly (2-Mark Questions)
1. State the principal value ranges of sin⁻¹ x, cos⁻¹ x, and tan⁻¹ x.
sin⁻¹ x has domain [-1, 1] and range [-pi/2, pi/2]. cos⁻¹ x has domain [-1, 1] and range [0, pi]. tan⁻¹ x has domain all real numbers and range (-pi/2, pi/2). The principal value is the unique value of the inverse function within its specified range. These ranges are chosen so that each inverse function is one-to-one and well-defined.
2. Prove that sin⁻¹ x + cos⁻¹ x = pi/2 for x in [-1, 1].
Let sin⁻¹ x = theta, so sin(theta) = x and theta is in [-pi/2, pi/2]. Then cos(pi/2 - theta) = sin(theta) = x. Since pi/2 - theta is in [0, pi] (which is the range of cos⁻¹), we have cos⁻¹ x = pi/2 - theta. Therefore sin⁻¹ x + cos⁻¹ x = theta + (pi/2 - theta) = pi/2. This identity holds for every x in the domain [-1, 1].
3. Evaluate cos⁻¹(cos 4pi/3).
Since 4pi/3 is not in [0, pi], we cannot directly write cos⁻¹(cos 4pi/3) = 4pi/3. Instead, cos(4pi/3) = cos(2pi - 4pi/3) = cos(2pi/3). Since 2pi/3 is in [0, pi], we get cos⁻¹(cos 4pi/3) = 2pi/3. This demonstrates the importance of checking whether the angle lies within the principal range.
4. Find the principal value of tan⁻¹(sqrt(3)).
We need to find theta in (-pi/2, pi/2) such that tan(theta) = sqrt(3). We know tan(pi/3) = sqrt(3) and pi/3 lies in (-pi/2, pi/2). Therefore, tan⁻¹(sqrt(3)) = pi/3. The principal value is pi/3 radians or 60 degrees.
5. Show that tan⁻¹ x + cot⁻¹ x = pi/2 for all real x.
Let tan⁻¹ x = theta, so tan(theta) = x and theta is in (-pi/2, pi/2). Then cot(pi/2 - theta) = tan(theta) = x. Since pi/2 - theta is in (0, pi) (the range of cot⁻¹), we have cot⁻¹ x = pi/2 - theta. Adding: tan⁻¹ x + cot⁻¹ x = theta + pi/2 - theta = pi/2. This identity holds for all real x.
VI. Answer in Detail (5-Mark Questions)
1. Explain the concept of principal values of inverse trigonometric functions and evaluate sin⁻¹(sin 3pi/5), cos⁻¹(cos 7pi/6), and tan⁻¹(tan 3pi/4).
The principal value is the unique angle that an inverse trigonometric function returns, which must lie within the defined principal range of that function. For sin⁻¹, the range is [-pi/2, pi/2]; for cos⁻¹, it is [0, pi]; for tan⁻¹, it is (-pi/2, pi/2). When evaluating sin⁻¹(sin theta), if theta is already in the principal range, the answer is simply theta. Otherwise, we must find an equivalent angle within the range that gives the same sine value. For sin⁻¹(sin 3pi/5): since 3pi/5 > pi/2, it is outside the range. sin(3pi/5) = sin(pi - 3pi/5) = sin(2pi/5). Since 2pi/5 is in [-pi/2, pi/2], sin⁻¹(sin 3pi/5) = 2pi/5. For cos⁻¹(cos 7pi/6): since 7pi/6 > pi, it is outside [0, pi]. cos(7pi/6) = cos(2pi - 7pi/6) = cos(5pi/6). Since 5pi/6 is in [0, pi], cos⁻¹(cos 7pi/6) = 5pi/6. For tan⁻¹(tan 3pi/4): since 3pi/4 > pi/2, it is outside (-pi/2, pi/2). tan(3pi/4) = tan(3pi/4 - pi) = tan(-pi/4). Since -pi/4 is in (-pi/2, pi/2), tan⁻¹(tan 3pi/4) = -pi/4.
2. Derive and explain the addition and subtraction formulas for tan⁻¹. Solve tan⁻¹(2) + tan⁻¹(3) using the appropriate formula.
The addition formula for tan⁻¹ states: tan⁻¹ x + tan⁻¹ y = tan⁻¹((x + y)/(1 - xy)) if xy < 1, and tan⁻¹ x + tan⁻¹ y = pi + tan⁻¹((x + y)/(1 - xy)) if xy > 1 and x, y > 0. Derivation: let alpha = tan⁻¹ x and beta = tan⁻¹ y. Then tan(alpha) = x and tan(beta) = y. Now tan(alpha + beta) = (tan alpha + tan beta)/(1 - tan alpha tan beta) = (x + y)/(1 - xy). So alpha + beta = tan⁻¹((x + y)/(1 - xy)), provided the result is in (-pi/2, pi/2). When xy > 1 with x, y > 0, the result exceeds pi/2, so we add pi to bring it into range. The subtraction formula is: tan⁻¹ x - tan⁻¹ y = tan⁻¹((x - y)/(1 + xy)). For tan⁻¹(2) + tan⁻¹(3): since xy = 6 > 1 and both x, y > 0, we use tan⁻¹(2) + tan⁻¹(3) = pi + tan⁻¹((2 + 3)/(1 - 6)) = pi + tan⁻¹(5/(-5)) = pi + tan⁻¹(-1) = pi - pi/4 = 3pi/4.
| 3. **Prove the following identities: (i) 2 tan⁻¹ x = sin⁻¹(2x/(1+x²)) for | x | <= 1, (ii) 2 tan⁻¹ x = cos⁻¹((1-x²)/(1+x²)) for | x | <= 1.** |
|---|
Proof of (i): Let tan⁻¹ x = theta, so tan(theta) = x. Then 2 tan⁻¹ x = 2 theta. We need to show sin(2 theta) = 2x/(1 + x²). Now sin(2 theta) = 2 sin(theta) cos(theta) = 2 tan(theta) cos²(theta) = 2x cos²(theta). Since tan(theta) = x, we have cos²(theta) = 1/(1 + tan²(theta)) = 1/(1 + x²). Therefore sin(2 theta) = 2x/(1 + x²), proving that 2 tan⁻¹ x = sin⁻¹(2x/(1 + x²)). Proof of (ii): Similarly, cos(2 theta) = cos²(theta) - sin²(theta) = (1 - tan²(theta))/(1 + tan²(theta)) = (1 - x²)/(1 + x²). Therefore 2 tan⁻¹ x = cos⁻¹((1 - x²)/(1 + x²)). These identities are useful for converting between inverse trigonometric forms and simplifying expressions in board exams.
4. Explain how to evaluate sin⁻¹(sin theta) when theta is outside the principal range. Give three examples.
When theta lies within the principal range [-pi/2, pi/2] for sin⁻¹, we simply write sin⁻¹(sin theta) = theta. When theta is outside this range, we use the property sin(pi - theta) = sin(theta) to find an equivalent angle within the range. Specifically: if theta is in (pi/2, pi], then sin⁻¹(sin theta) = pi - theta (which lies in [0, pi/2]). If theta is in (pi, 3pi/2], then sin⁻¹(sin theta) = theta - pi (which lies in (-pi/2, 0] but we adjust). Example 1: sin⁻¹(sin 2pi/3). Since 2pi/3 is in (pi/2, pi], sin(2pi/3) = sin(pi - 2pi/3) = sin(pi/3). So sin⁻¹(sin 2pi/3) = pi/3. Example 2: sin⁻¹(sin 5pi/4). Since 5pi/4 is in (pi, 3pi/2], sin(5pi/4) = -sin(pi/4) = -1/sqrt(2). So sin⁻¹(sin 5pi/4) = sin⁻¹(-1/sqrt(2)) = -pi/4. Example 3: sin⁻¹(sin 7pi/6). Since 7pi/6 is in (pi, 3pi/2], sin(7pi/6) = -sin(pi/6) = -1/2. So sin⁻¹(sin 7pi/6) = sin⁻¹(-1/2) = -pi/6.
Common Mistakes to Avoid
1. ❌ Writing sin⁻¹(sin theta) = theta for all values of theta. ✅ sin⁻¹(sin theta) = theta only when theta is in [-pi/2, pi/2]. Outside this range, adjust the angle.
2. ❌ Confusing the principal ranges of different inverse trigonometric functions. ✅ Memorise: sin⁻¹ in [-pi/2, pi/2], cos⁻¹ in [0, pi], tan⁻¹ in (-pi/2, pi/2), cot⁻¹ in (0, pi).
3. ❌ Applying the addition formula tan⁻¹ x + tan⁻¹ y = tan⁻¹((x+y)/(1-xy)) without checking the condition xy < 1. ✅ When xy > 1 and x, y > 0, add pi to the result: tan⁻¹ x + tan⁻¹ y = pi + tan⁻¹((x+y)/(1-xy)).
4. ❌ Writing degree values when the question expects radians. ✅ Always express inverse trigonometric values in radians unless degrees are explicitly asked.
5. ❌ Ignoring domain restrictions when composing inverse and direct functions. ✅ sin⁻¹(2) is undefined because 2 is outside [-1, 1]. Always check the domain before computing.
6. ❌ Assuming cos⁻¹(cos theta) = theta for all theta. ✅ cos⁻¹(cos theta) = theta only when theta is in [0, pi]. Outside this range, find the equivalent angle in [0, pi].
7. ❌ Forgetting that sin⁻¹(-x) = -sin⁻¹(x) (odd function) but cos⁻¹(-x) = pi - cos⁻¹(x) (not simply negative). ✅ Use the correct sign property: sin⁻¹ is odd, but cos⁻¹(-x) = pi - cos⁻¹(x).
8. ❌ Substituting numerical values directly without checking the applicable range first. ✅ Always identify which formula applies (addition, subtraction, or complementary) based on the sign and magnitude of the arguments.
Exam Checklist
- Check the principal value range before writing the final answer for any inverse trigonometric function.
- Write the relevant identity before substituting values in simplification problems.
- Verify that sin⁻¹(sin theta) answers lie in [-pi/2, pi/2] and cos⁻¹(cos theta) answers lie in [0, pi].
- Use the addition formula only after verifying xy < 1 or xy > 1 to decide whether pi needs to be added.
- Express final answers in radians unless the question specifies degrees.
- Check domain restrictions: sin⁻¹ and cos⁻¹ require arguments in [-1, 1]; sec⁻¹ and cosec⁻¹ require |x| >= 1.
- For double-angle inverse formulas, verify that the intermediate results stay within the principal range.
- Draw a quick sketch of the unit circle to verify quadrant-based answers.
- Show all intermediate steps, especially the adjustment of angles outside the principal range.
- Cross-verify answers by taking the direct trigonometric function of the result and checking if it matches the original argument.