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Inverse Trigonometric Functions: JEE Main Previous Year Questions (2021-2026)
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Page 1 JEE Main Previous Year Questions (2021-2026): Inverse Trigonometric Functions (January 2026) Q1: Considering the principal values of inverse trigonometric functions, the value of the expression is equal to: A: 33/56 B: C: D: 16/63 Answer: A Explanation: Simplify the individual components Let a = sin ?¹(2/v13) and ß = cos ?¹(3/v10). Page 2 JEE Main Previous Year Questions (2021-2026): Inverse Trigonometric Functions (January 2026) Q1: Considering the principal values of inverse trigonometric functions, the value of the expression is equal to: A: 33/56 B: C: D: 16/63 Answer: A Explanation: Simplify the individual components Let a = sin ?¹(2/v13) and ß = cos ?¹(3/v10). Q2: If the domain of the function f(x) = sin ?¹(1/(x²-2x-2)), is (-8, a] ? [ß, ?] ? [d, 8), then a + ß + ? + d is equal to A: 4 B: 2 C: 5 D: 3 Answer: A Explanation: Page 3 JEE Main Previous Year Questions (2021-2026): Inverse Trigonometric Functions (January 2026) Q1: Considering the principal values of inverse trigonometric functions, the value of the expression is equal to: A: 33/56 B: C: D: 16/63 Answer: A Explanation: Simplify the individual components Let a = sin ?¹(2/v13) and ß = cos ?¹(3/v10). Q2: If the domain of the function f(x) = sin ?¹(1/(x²-2x-2)), is (-8, a] ? [ß, ?] ? [d, 8), then a + ß + ? + d is equal to A: 4 B: 2 C: 5 D: 3 Answer: A Explanation: so these two inequality and final solution will be intersection of solution of these two inequality. Page 4 JEE Main Previous Year Questions (2021-2026): Inverse Trigonometric Functions (January 2026) Q1: Considering the principal values of inverse trigonometric functions, the value of the expression is equal to: A: 33/56 B: C: D: 16/63 Answer: A Explanation: Simplify the individual components Let a = sin ?¹(2/v13) and ß = cos ?¹(3/v10). Q2: If the domain of the function f(x) = sin ?¹(1/(x²-2x-2)), is (-8, a] ? [ß, ?] ? [d, 8), then a + ß + ? + d is equal to A: 4 B: 2 C: 5 D: 3 Answer: A Explanation: so these two inequality and final solution will be intersection of solution of these two inequality. x ? (-8, -1] ? (1 - v3, 1 + v3) ? [3, 8) - S 2 taking intersection of S 1 & S 2 for final solution. so domain of f(x) is x ? (-8, -1] ? [1 - v2, 1 + v2] ? [3, 8) compare this with given domain x ? (-8, a] ? [ß, ?] ? [d, 8) so, a = -1, ß = 1 - v2, ? = 1 + v2, d = 3 a + ß + ? + d = -1 + 1 - v2 + 1 + v2 + 3 = 4 Q3: The number of solutions of tan ?¹ 4x + tan ?¹ 6x = p/6, where is equal to: A: 2 B: 0 C: 3 D: 1 Answer: D Explanation: Page 5 JEE Main Previous Year Questions (2021-2026): Inverse Trigonometric Functions (January 2026) Q1: Considering the principal values of inverse trigonometric functions, the value of the expression is equal to: A: 33/56 B: C: D: 16/63 Answer: A Explanation: Simplify the individual components Let a = sin ?¹(2/v13) and ß = cos ?¹(3/v10). Q2: If the domain of the function f(x) = sin ?¹(1/(x²-2x-2)), is (-8, a] ? [ß, ?] ? [d, 8), then a + ß + ? + d is equal to A: 4 B: 2 C: 5 D: 3 Answer: A Explanation: so these two inequality and final solution will be intersection of solution of these two inequality. x ? (-8, -1] ? (1 - v3, 1 + v3) ? [3, 8) - S 2 taking intersection of S 1 & S 2 for final solution. so domain of f(x) is x ? (-8, -1] ? [1 - v2, 1 + v2] ? [3, 8) compare this with given domain x ? (-8, a] ? [ß, ?] ? [d, 8) so, a = -1, ß = 1 - v2, ? = 1 + v2, d = 3 a + ß + ? + d = -1 + 1 - v2 + 1 + v2 + 3 = 4 Q3: The number of solutions of tan ?¹ 4x + tan ?¹ 6x = p/6, where is equal to: A: 2 B: 0 C: 3 D: 1 Answer: D Explanation: Q4: If the domain of the function is the interval [a, ß], then a + 2ß is equal to: A: 5 B: 2 C: 3 D: 1 Answer: C Explanation:Read More
FAQs on Inverse Trigonometric Functions: JEE Main Previous Year Questions (2021-2026)
| 1. What are the basic properties of inverse trigonometric functions? |  |
Ans. The basic properties of inverse trigonometric functions include the principal value branches, range, domain, and periodicity of functions such as sin$^{-1}x$, cos$^{-1}x$, and tan$^{-1}x$.
| 2. How do we find the derivatives of inverse trigonometric functions? | |
Ans. To find the derivatives of inverse trigonometric functions, we can use the chain rule and the known derivatives of trigonometric functions. For example, the derivative of sin$^{-1}x$ is $\frac{1}{\sqrt{1-x^2}}$.
| 3. What are some common identities involving inverse trigonometric functions? | |
Ans. Some common identities involving inverse trigonometric functions include sin(sin$^{-1}x) = x$, cos(cos$^{-1}x) = x$, and tan(tan$^{-1}x) = x$. These identities help simplify expressions involving inverse trigonometric functions.
| 4. How do we solve equations involving inverse trigonometric functions? | |
Ans. To solve equations involving inverse trigonometric functions, we can apply the properties and identities of these functions. We can also use trigonometric identities to simplify the expressions and find the solutions.
| 5. Can inverse trigonometric functions be used to find angles in a triangle? | |
Ans. Yes, inverse trigonometric functions can be used to find angles in a triangle. By using the properties of inverse trigonometric functions, we can determine unknown angles in a triangle based on the given sides or angles.
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