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Inverse Trigonometric Functions: JEE Main Previous Year Questions (2021-2026)

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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: 
 
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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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