Inverse Trigonometric Functions: Exercise 2.2 (Class 12 Maths NCERT)

Understanding Exercise 2.2: The Core Idea

Exercise 2.2 of your NCERT textbook primarily focuses on applying various properties of inverse trigonometric functions to either simplify expressions or prove specific identities. Unlike the previous exercise, which largely dealt with principal value branches, this section challenges you to manipulate and transform expressions. The key strategy here is to convert inverse trigonometric functions into standard trigonometric forms using suitable substitutions or directly apply established identities. For instance, an expression like sin⁻¹(2x√(1-x²)) immediately suggests a substitution x = sin θ because 2sinθcosθ = sin(2θ). Similarly, problems involving tan⁻¹ often require the sum/difference formulas for tan⁻¹x or the 2tan⁻¹x identities. A strong foundation in basic trigonometric identities from Class 11 is indispensable here, as inverse trigonometric properties are often derived from them. Pay close attention to the domain and range of each inverse function when making substitutions or applying identities, as these restrictions can sometimes alter the final form or require careful adjustments.

Essential Properties for Solving Exercise 2.2

Identities with 2tan⁻¹x

These are fundamental for simplification and proofs: 1. 2tan⁻¹x = sin⁻¹(2x / (1 + x²)) for |x| ≤ 1 2. 2tan⁻¹x = cos⁻¹((1 - x²) / (1 + x²)) for x ≥ 0 3. 2tan⁻¹x = tan⁻¹(2x / (1 - x²)) for -1 < x < 1 These allow interconversion between tan⁻¹, sin⁻¹, and cos⁻¹.

Sum and Difference Identities

Crucial for combining or expanding inverse tangent functions: 1. tan⁻¹x + tan⁻¹y = tan⁻¹((x + y) / (1 - xy)) for xy < 1 2. tan⁻¹x - tan⁻¹y = tan⁻¹((x - y) / (1 + xy)) for xy > -1 There are also variations for xy > 1 (adding or subtracting π), which are less common but important to remember for specific cases.

Substitution-Based Simplification

Many problems require substituting x = sin θ, x = cos θ, x = tan θ, x = a sin θ, x = a tan θ, etc., to convert the inverse trigonometric expression into a standard trigonometric identity, which can then be simplified. For example, for √(1-x²), substitute x = sin θ or x = cos θ.

Identities for 3sin⁻¹x, 3cos⁻¹x, 3tan⁻¹x

These are direct applications of multiple angle formulas from trigonometry: 1. 3sin⁻¹x = sin⁻¹(3x - 4x³) for x ∈ [-1/2, 1/2] 2. 3cos⁻¹x = cos⁻¹(4x³ - 3x) for x ∈ [1/2, 1] 3. 3tan⁻¹x = tan⁻¹((3x - x³) / (1 - 3x²)) for x ∈ (-1/√3, 1/√3)

Fully Worked Examples from Exercise 2.2

• Example 1: Simplify the expression tan⁻¹((3a²x - x³)/(a³ - 3ax²)) where a > 0 and -a/√3 < x < a/√3. Step 1: Observe the structure of the expression. It resembles the formula for tan(3θ) = (3tanθ - tan³θ) / (1 - 3tan²θ). This suggests a substitution involving tan θ. Step 2: Let x = a tan θ. Then tan θ = x/a. Since -a/√3 < x < a/√3, we have -1/√3 < x/a < 1/√3, which implies -π/6 < θ < π/6. Step 3: Substitute x = a tan θ into the expression: tan⁻¹((3a²(a tan θ) - (a tan θ)³) / (a³ - 3a(a tan θ)²)) = tan⁻¹((3a³ tan θ - a³ tan³ θ) / (a³ - 3a³ tan² θ)) Step 4: Factor out a³ from both the numerator and denominator: = tan⁻¹(a³(3 tan θ - tan³ θ) / a³(1 - 3 tan² θ)) = tan⁻¹((3 tan θ - tan³ θ) / (1 - 3 tan² θ)) Step 5: Apply the trigonometric identity tan(3θ) = (3 tan θ - tan³ θ) / (1 - 3 tan² θ): = tan⁻¹(tan(3θ)) Step 6: Since -π/6 < θ < π/6, it follows that -π/2 < 3θ < π/2. In this interval, tan⁻¹(tan y) = y. Therefore: = 3θ Step 7: Substitute back θ = tan⁻¹(x/a): = 3tan⁻¹(x/a) Final answer: 3tan⁻¹(x/a)
• Example 2: Prove that tan⁻¹(2/11) + tan⁻¹(7/24) = tan⁻¹(1/2). Step 1: Use the tan⁻¹x + tan⁻¹y identity. Here, x = 2/11 and y = 7/24. Step 2: Check the condition xy < 1: xy = (2/11) (7/24) = 14 / 264 = 7 / 132. Since 7/132 < 1, we can directly apply the formula: tan⁻¹x + tan⁻¹y = tan⁻¹((x + y) / (1 - xy)) Step 3: Substitute the values of x and y: LHS = tan⁻¹((2/11 + 7/24) / (1 - (2/11)(7/24))) Step 4: Calculate the numerator: 2/11 + 7/24 = (224 + 711) / (11*24) = (48 + 77) / 264 = 125 / 264 Step 5: Calculate the denominator: 1 - (14/264) = (264 - 14) / 264 = 250 / 264 Step 6: Substitute these back into the expression: LHS = tan⁻¹((125/264) / (250/264)) = tan⁻¹(125 / 250) = tan⁻¹(1/2) Step 7: Compare with the RHS. LHS = tan⁻¹(1/2) = RHS Final answer: The identity is proven.
• Example 3: Write tan⁻¹((cos x - sin x) / (cos x + sin x)) in the simplest form, where -π/4 < x < 3π/4. Step 1: The expression involves sin x and cos x in a fractional form. A common strategy for such expressions inside tan⁻¹ is to divide the numerator and denominator by cos x. Step 2: Divide both numerator and denominator by cos x: tan⁻¹(((cos x - sin x) / cos x) / ((cos x + sin x) / cos x)) = tan⁻¹((1 - (sin x / cos x)) / (1 + (sin x / cos x))) = tan⁻¹((1 - tan x) / (1 + tan x)) Step 3: Recognize the form (1 - tan x) / (1 + tan x). This is related to the tangent subtraction formula tan(A - B) = (tan A - tan B) / (1 + tan A tan B). Here, tan(π/4) = 1. Step 4: Substitute 1 with tan(π/4): = tan⁻¹((tan(π/4) - tan x) / (1 + tan(π/4) tan x)) Step 5: Apply the identity tan(A - B) = (tan A - tan B) / (1 + tan A tan B): = tan⁻¹(tan(π/4 - x)) Step 6: Consider the given domain: -π/4 < x < 3π/4. We need to find the range of π/4 - x. Multiply by -1: -3π/4 < -x < π/4 Add π/4: π/4 - 3π/4 < π/4 - x < π/4 + π/4 -π/2 < π/4 - x < π/2 Since π/4 - x lies in the principal value branch (-π/2, π/2) of tan⁻¹, we can write tan⁻¹(tan(y)) = y. Step 7: Therefore, the simplified form is: = π/4 - x Final answer: π/4 - x

Mastering Exercise 2.2: Exam Tips

To excel in problems from Exercise 2.2, keep these critical points in mind:

1. Know Your Substitutions: The trickiest part is often choosing the right substitution. If you see √(1-x²), think x = sin θ or x = cos θ. For √(1+x²), try x = tan θ or x = cot θ. For √(x²-1), consider x = sec θ or x = cosec θ. Expressions like (1-x²)/(1+x²) often suggest x = tan θ to simplify into cos(2θ). Practice recognizing these patterns.

1. Domain and Range Matters: Always pay attention to the given conditions for x (e.g., |x| < 1, x ≥ 0). These conditions are crucial for determining the correct range of θ after substitution, which in turn ensures that sin⁻¹(sin θ) = θ or tan⁻¹(tan θ) = θ correctly. If θ falls outside the principal value branch, you'll need to adjust using properties like sin⁻¹(sin θ) = π - θ (if θ is in [π/2, 3π/2]). Ignoring this is a common error that leads to incorrect answers.

1. Identities, Identities, Identities: Memorize the key identities for 2tan⁻¹x (in terms of sin⁻¹, cos⁻¹, and tan⁻¹) and the sum/difference formulas for tan⁻¹x. These are frequently tested. Also, be familiar with the 3sin⁻¹x, 3cos⁻¹x, and 3tan⁻¹x identities as they often appear as direct proofs.

1. Simplify Trigonometric Expressions First: Before applying inverse function properties, simplify the inner trigonometric expression as much as possible using your Class 11 identities. For example, (1 - cos x) / sin x simplifies to tan(x/2).

1. Practice Proving Identities: When proving LHS = RHS, either start from one side and reach the other, or simplify both sides to a common expression. Clearly state the identity or substitution you are using at each step.

Practice Questions with Solutions

• Q: Prove that 3cos⁻¹x = cos⁻¹(4x³ - 3x) for x ∈ [1/2, 1]. A: Step 1: Let x = cos θ. Since x ∈ [1/2, 1], cos θ ∈ [1/2, 1]. This implies θ ∈ [0, π/3]. Step 2: Substitute x = cos θ into the RHS: RHS = cos⁻¹(4cos³θ - 3cosθ) Step 3: Recognize the trigonometric identity cos(3θ) = 4cos³θ - 3cosθ: RHS = cos⁻¹(cos(3θ)) Step 4: Check if 3θ lies in the principal value branch of cos⁻¹x, which is [0, π]. Since θ ∈ [0, π/3], then 3θ ∈ [0, π]. Thus, cos⁻¹(cos(3θ)) = 3θ. Step 5: Substitute back θ = cos⁻¹x: RHS = 3cos⁻¹x Step 6: This is equal to the LHS. Final answer: Proven.
• Q: Simplify tan⁻¹((√(1+x²) - 1) / x), where x ≠ 0. A: Step 1: Observe the form √(1+x²). This suggests the substitution x = tan θ. Then θ = tan⁻¹x. Step 2: Substitute x = tan θ into the expression: tan⁻¹((√(1+tan²θ) - 1) / tan θ) Step 3: Use the identity 1+tan²θ = sec²θ: tan⁻¹((√(sec²θ) - 1) / tan θ) = tan⁻¹((|sec θ| - 1) / tan θ) Assuming θ ∈ (-π/2, π/2), sec θ is positive, so |sec θ| = sec θ. = tan⁻¹((sec θ - 1) / tan θ) Step 4: Convert sec θ and tan θ to sin θ and cos θ: = tan⁻¹(((1/cos θ) - 1) / (sin θ / cos θ)) = tan⁻¹(((1 - cos θ) / cos θ) / (sin θ / cos θ)) = tan⁻¹((1 - cos θ) / sin θ) Step 5: Use half-angle identities: 1 - cos θ = 2sin²(θ/2) and sin θ = 2sin(θ/2)cos(θ/2): = tan⁻¹((2sin²(θ/2)) / (2sin(θ/2)cos(θ/2))) = tan⁻¹(sin(θ/2) / cos(θ/2)) = tan⁻¹(tan(θ/2)) Step 6: Since x = tan θ and x ≠ 0, θ ≠ 0. For θ = tan⁻¹x, we know θ ∈ (-π/2, π/2). Therefore, θ/2 ∈ (-π/4, π/4). In this range, tan⁻¹(tan(y)) = y. = θ/2 Step 7: Substitute back θ = tan⁻¹x: = (1/2)tan⁻¹x Final answer: (1/2)tan⁻¹x
• Q: Find the value of tan( (1/2)sin⁻¹(2x/(1+x²)) + (1/2)cos⁻¹((1-y²)/(1+y²)) ), for |x| < 1, y > 0 and xy < 1. A: Step 1: Recognize the standard identities for 2tan⁻¹x. We know sin⁻¹(2x/(1+x²)) = 2tan⁻¹x for |x| ≤ 1. And cos⁻¹((1-y²)/(1+y²)) = 2tan⁻¹y for y ≥ 0. Step 2: Substitute these into the expression: tan( (1/2)(2tan⁻¹x) + (1/2)(2tan⁻¹y) ) = tan(tan⁻¹x + tan⁻¹y) Step 3: Apply the tan⁻¹x + tan⁻¹y formula. Given xy < 1: = tan(tan⁻¹((x + y) / (1 - xy))) Step 4: Use the property tan(tan⁻¹A) = A: = (x + y) / (1 - xy) Final answer: (x + y) / (1 - xy)
• Q: Prove that tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5). A: Step 1: Let sin⁻¹(5/13) = A. Then sin A = 5/13. We need tan A. Using cos²A = 1 - sin²A, cos A = √(1 - (5/13)²) = √(1 - 25/169) = √(144/169) = 12/13 (since A is typically in (-π/2, π/2), cos A is positive). So tan A = sin A / cos A = (5/13) / (12/13) = 5/12. Step 2: Let cos⁻¹(3/5) = B. Then cos B = 3/5. We need tan B. Using sin²B = 1 - cos²B, sin B = √(1 - (3/5)²) = √(1 - 9/25) = √(16/25) = 4/5 (since B is typically in [0, π], sin B is positive). So tan B = sin B / cos B = (4/5) / (3/5) = 4/3. Step 3: Now the RHS becomes A + B = tan⁻¹(5/12) + tan⁻¹(4/3). Step 4: Apply the tan⁻¹x + tan⁻¹y formula. Check xy: (5/12) (4/3) = 20/36 = 5/9. Since 5/9 < 1, we use tan⁻¹((x+y)/(1-xy)). RHS = tan⁻¹(((5/12) + (4/3)) / (1 - (5/12)(4/3))) = tan⁻¹(((15+48)/36) / (1 - 20/36)) = tan⁻¹((63/36) / ((36-20)/36)) = tan⁻¹((63/36) / (16/36)) = tan⁻¹(63/16) Step 5: This is equal to the LHS. Final answer: Proven.

Frequently Asked Questions