Inverse Trigonometric Functions Miscellaneous Ex | Class 12 NCERT

Decoding the Miscellaneous Exercise: What to Expect

The Miscellaneous Exercise for Inverse Trigonometric Functions is designed to push you beyond simple formula application. The problems are often multi-layered and test your understanding in a comprehensive way. You'll find that many questions require you to first simplify a complex expression inside the inverse function before you can even begin to solve. The key is to recognize patterns. For instance, an expression like 2x / (1 + x²) should immediately remind you of the tan 2θ formula if you substitute x = tan θ, leading to sin⁻¹(sin 2θ). Similarly, solving equations isn't just about finding a value for 'x'. It involves using properties like tan⁻¹A + tan⁻¹B, simplifying, and then solving the resulting algebraic equation. Most importantly, every problem demands a constant awareness of the principal value branches of the six inverse trigonometric functions. A solution that is algebraically correct might be mathematically incorrect if it falls outside the defined range.

A Strategic Approach to Solving Problems

1. Step 1: Analyze and Identify the Goal — First, read the question carefully. Is it a 'Prove That' question, a 'Solve for x' equation, or a 'Find the value of' simplification? This determines your overall strategy. Identify the inverse functions involved and the structure of the argument.
2. Step 2: Look for Simplification via Properties — Before any complex steps, see if you can apply standard properties. Can you use tan⁻¹x + tan⁻¹y? Or 2tan⁻¹x? Can you convert one inverse function to another (e.g., sin⁻¹ to tan⁻¹) to make the problem uniform? For example, to prove sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36), it's best to convert both sin⁻¹ terms to tan⁻¹ first and then apply the tan⁻¹x + tan⁻¹y formula.
3. Step 3: Use Trigonometric Substitutions — If you see algebraic expressions involving x, a trigonometric substitution is often the key. Common substitutions are: - For √(a² - x²), use x = a sin θ or x = a cos θ. - For a² + x² or √(a² + x²), use x = a tan θ or x = a cot θ. - For x² - a² or √(x² - a²), use x = a sec θ or x = a cosec θ. This converts the algebraic expression into a simpler trigonometric one.
4. Step 4: Solve and Verify in Principal Branch — After simplifying, solve the equation or complete the proof. The final and most crucial step is verification. Ensure your final answer for 'x' is valid within the domain. For simplification problems, ensure the final angle lies within the principal value branch of the respective inverse function. For example, the answer for sin⁻¹(y) must be in [-π/2, π/2].

Worked Examples from the Miscellaneous Exercise

• Problem 1: Solve the equation tan⁻¹((x-1)/(x-2)) + tan⁻¹((x+1)/(x+2)) = π/4 Step 1: Apply the tan⁻¹A + tan⁻¹B formula. The formula is tan⁻¹A + tan⁻¹B = tan⁻¹((A+B)/(1-AB)). Here, A = (x-1)/(x-2) and B = (x+1)/(x+2). So, tan⁻¹[ { (x-1)/(x-2) + (x+1)/(x+2) } / { 1 - ((x-1)/(x-2)) * ((x+1)/(x+2)) } ] = π/4 Step 2: Simplify the expression inside tan⁻¹. Numerator = ((x-1)(x+2) + (x+1)(x-2)) / ((x-2)(x+2)) = (x²+x-2 + x²-x-2) / (x²-4) = (2x²-4) / (x²-4) Denominator = 1 - (x²-1)/(x²-4) = ((x²-4) - (x²-1)) / (x²-4) = (-3) / (x²-4) The expression becomes tan⁻¹( (2x²-4)/(-3) ) = π/4 Step 3: Solve for x. Taking tan on both sides: (2x²-4)/(-3) = tan(π/4) (2x²-4)/(-3) = 1 2x² - 4 = -3 2x² = 1 x² = 1/2 x = ±1/√2 Final Answer: The solutions are x = 1/√2 and x = -1/√2.
• Problem 2: Prove that 9π/8 - (9/4)sin⁻¹(1/3) = (9/4)sin⁻¹(2√2/3) Step 1: Rearrange the equation and simplify. Start with the Left Hand Side (LHS): 9π/8 - (9/4)sin⁻¹(1/3) Take (9/4) common: (9/4) [ π/2 - sin⁻¹(1/3) ] Step 2: Apply the identity sin⁻¹x + cos⁻¹x = π/2. We know that cos⁻¹x = π/2 - sin⁻¹x. Using this, we can replace [ π/2 - sin⁻¹(1/3) ] with cos⁻¹(1/3). So, the LHS becomes (9/4)cos⁻¹(1/3). Step 3: Convert cos⁻¹ to sin⁻¹. Let cos⁻¹(1/3) = θ. This means cos θ = 1/3. (Base/Hypotenuse) We need to find sin θ. Using Pythagoras theorem, Perpendicular = √(Hypotenuse² - Base²) = √(3² - 1²) = √8 = 2√2. So, sin θ = Perpendicular/Hypotenuse = (2√2)/3. This means θ = sin⁻¹(2√2/3). Therefore, cos⁻¹(1/3) = sin⁻¹(2√2/3). Step 4: Substitute back into the expression. Our expression was (9/4)cos⁻¹(1/3). Substituting the converted value, we get (9/4)sin⁻¹(2√2/3). This is equal to the Right Hand Side (RHS). Final Answer: Hence, LHS = RHS, and the identity is proved.

Exam Trap: Principal Values and Domain

A very common mistake students make in the board exam is ignoring the principal value branches. For example, if you get an expression like sin⁻¹(sin(3)), the answer is not 3. Remember, the range of sin⁻¹x is [-π/2, π/2], which is approximately [-1.57, 1.57]. The value 3 lies outside this range. You must find an equivalent angle within the range. We know sin(π - x) = sin(x). So, sin(3) = sin(π - 3). Since π - 3 (approx 3.14 - 3 = 0.14) lies within [-π/2, π/2], the correct answer is π - 3. Always double-check that your final simplified value lies within the principal value branch of the outermost function. Failing to do this can lead to a loss of significant marks, even if your algebraic steps were correct.

Practice Questions with Solutions

• Q: Find the value of tan(1/2 [sin⁻¹(2x/(1+x²)) + cos⁻¹((1-y²)/(1+y²))]), where |x|<1, y>0 and xy<1. A: Step 1: Recognize the standard substitution formulas. The expressions inside are standard formulas if we substitute x = tan A and y = tan B. sin⁻¹(2x/(1+x²)) = sin⁻¹(2tanA/(1+tan²A)) = sin⁻¹(sin 2A) = 2A = 2tan⁻¹x. cos⁻¹((1-y²)/(1+y²)) = cos⁻¹((1-tan²B)/(1+tan²B)) = cos⁻¹(cos 2B) = 2B = 2tan⁻¹y. Step 2: Substitute these simplified forms back into the original expression. tan(1/2 [2tan⁻¹x + 2tan⁻¹y]) = tan([tan⁻¹x + tan⁻¹y]) Step 3: Apply the tan⁻¹x + tan⁻¹y formula. = tan(tan⁻¹((x+y)/(1-xy))) Step 4: Simplify the final expression. = (x+y)/(1-xy) Final answer: (x+y)/(1-xy)
• Q: Solve for x: 2tan⁻¹(cos x) = tan⁻¹(2 cosec x) A: Step 1: Apply the 2tan⁻¹A formula on the LHS. The formula is 2tan⁻¹A = tan⁻¹(2A / (1-A²)). Here, A = cos x. LHS = tan⁻¹(2cos x / (1-cos²x)) = tan⁻¹(2cos x / sin²x). Step 2: Equate the LHS and RHS. tan⁻¹(2cos x / sin²x) = tan⁻¹(2 cosec x) (2cos x / sin²x) = 2 cosec x (2cos x / sin²x) = 2 / sin x Step 3: Solve the resulting trigonometric equation. Assuming sin x ≠ 0. cos x / sin x = 1 cot x = 1 Step 4: Find the principal value for x. cot x = 1 implies x = π/4. Final answer: x = π/4
• Q: Prove that tan⁻¹(√x) = (1/2)cos⁻¹((1-x)/(1+x)), x ∈ [0, 1]. A: Step 1: Start with the RHS of the equation. RHS = (1/2)cos⁻¹((1-x)/(1+x)) Step 2: Use a trigonometric substitution. Let x = tan²θ. This is a good choice because it simplifies the expression inside cos⁻¹. RHS = (1/2)cos⁻¹((1-tan²θ)/(1+tan²θ)) Step 3: Apply the formula for cos 2θ. We know cos 2θ = (1-tan²θ)/(1+tan²θ). RHS = (1/2)cos⁻¹(cos 2θ) = (1/2)(2θ) = θ Step 4: Substitute back to x. We assumed x = tan²θ, so √x = tan θ. This implies θ = tan⁻¹(√x). Since RHS simplified to θ, we have RHS = tan⁻¹(√x). This is equal to the LHS. Final answer: Hence, proved.
• Q: Find the value of sin⁻¹(sin(10)). A: Step 1: Check if the value is in the principal branch. The principal value branch for sin⁻¹ is [-π/2, π/2], which is approximately [-1.57, 1.57]. The value 10 is far outside this range. Step 2: Find an integer multiple of π close to 10. We know π ≈ 3.14. So, 3π ≈ 9.42. Step 3: Express 10 in terms of the nearest multiple of π. We can write 10 = 3π + (10 - 3π). Or we can express sin(10) differently. Let's use the identity sin(x) = sin(nπ ± x) depending on n. We know sin(x) = sin(x - 2kπ). We also know sin(x) = sin(π-x), sin(x) = sin(x+2π) etc. Let's find which quadrant 10 radians is in. 3π ≈ 9.42 and 3.5π ≈ 10.99. So 10 is in the 3rd quadrant. In the 3rd quadrant, sin is negative. We need an angle in [-π/2, π/2]. We can use the identity sin(x) = sin(x - nπ) with appropriate sign changes. A simpler way is to find a value y = 10 - k(2π) or y = k(π) - 10 that falls in the range. Consider 3π - 10. sin(3π - x) = sin(x). So sin(10) = sin(3π - 10). Let's check the value of 3π - 10. It is approx 9.42 - 10 = -0.58. This value lies in [-π/2, π/2]. Step 4: Simplify the expression. sin⁻¹(sin(10)) = sin⁻¹(sin(3π - 10)) = 3π - 10. Final answer: 3π - 10

Frequently Asked Questions