**Important Questions for Class 12 Maths Chapter 2** – Inverse Trigonometric functions are given here. The questions are framed as per the syllabus of the CBSE Board. All the concepts in class 12 chapter 2 are covered here. You can go through the given important questions with solutions, also try to solve the practice problems provided on our website. Class 12 maths is an easy and important subject. The students can easily secure the good marks in the board examination. Also, get the class 12 Maths important questions for all the chapters here at BYJU’S.

Go through the given solved problems. Here, the solutions for the class 12 Maths important questions is given in a step by step format, so that students can easily understand the flow of solving the problems. Before, going through the important questions, let us have a look at concepts covered in the class 12 Maths Chapter 2 – Inverse Trigonometric Functions.

**Also, check:**

- Important 4 Marks Questions for CBSE Class 12 Maths
- Important 6 marks Questions for CBSE Class 12 Maths

**What is Inverse Trigonometric Function?**

We know that trigonometric functions are the many to one functions. If the function is bijective, we can find the inverse of the function. This is known as the inverse trigonometric function. If “f” is the function, then the inverse of the function is given as f^{-1}. For example: Let y = f(x) = cos x, then the inverse of the given trigonometric function is defined by x = cos^{-1} y. In this chapter 2 inverse trigonometric function class 12 Maths, the detailed explanation for the domain and range of the inverse trigonometric functions is provided along with the properties.

## Class 12 Chapter 2 – Inverse Trigonometric Function Important Questions with Solutions

Go through the solved problems on class 12 Chapter 2 Inverse Trigonometric Functions, which are given below. The following important questions with solutions help the students to score good marks in the board examination.

**Q.1:** Determine the principal value of cos^{-1}( -1/2).

**Solution:**

Let us assume that, y = cos^{-1}( -1/2)

We can write this as:

cos y =- 1/2

cos y = cos (2π/3).

Thus, the Range of the principal value of cos^{-1} is [0, π ].

Therefore, the principal value of cos^{-1}( -1/2) is 2π /3.

**Q.2:** Find the value of cot (tan^{-1} α + cot^{-1} α).

**Solution: **

Given that: cot (tan^{-1} 𝛂 + cot^{-1} 𝛂)

= cot (𝝅/𝟐) (since, tan^{-1} x + cot^{-1} x = 𝜋/2)

= cot (180°/2) ( we know that cot 90° = 0 )

= cot (90°)

= 0

Therefore, the value of cot (tan^{-1} α + cot^{-1} α) is 0.

**Q.3:** The value of tan^{-1} √3 – sec^{-1}(–2) is equal to:

(A) π (B) – π/3 (C) π/3 (D) 2π/3

**Solution:**

Now, solve the first part of the expression: tan^{-1} √3

Let us take y = tan^{-1}√3

This can be written as:

tan y = √3

Now, use the trigonometry table to find the radian value

tan y = tan (π/3)

Thus, the range of principal value of tan^{-1} is (−π/2, π/2)

Therefore, the principal value of tan^{-1}√3 is π/3.

Now, solve the second part of the expression: sec^{-1}(–2)

Now, assume that y = sec^{-1} (–2)

sec y = -2

sec y = sec (2π/3)

We know that the principal value range of sec^{-1} is [0,π] – {π/2}

Therefore, the principal value of sec^{-1} (–2) = 2π/3

Now we have:

tan^{-1}(√3) = π/3

sec^{-1} (–2) = 2π/3

Now, substitute the values in the given expression:

= tan^{-1} √3 – sec^{-1} (−2)

= π/3 − (2π/3)

= π/3 − 2π/3

= (π − 2π)/3

= – π/3

Hence, the correct answer is an option (B)

**Q.4:** Prove that sin^{-1} (3/5) – sin^{-1} (8/17) = cos^{-1} (84/85).

**Solution:**

Let sin^{-1} (3/5) = a and sin^{-1} (8/17) = b

Thus, we can write sin a = 3/5 and sin b = 8/17

Now, find the value of cos a and cos b

**To find cos a:**

Cos a = √[1 – sin^{2} a]

= √[1 – (3/5)^{2} ]

= √[1 – (9/25)]

= √[(25-9)/25]

= 4/5

Thus, the value of cos a = 4/5

**To find cos b:**

Cos b= √[1 – sin^{2} b]

= √[1 – (8/17)^{2} ]

= √[1 – (64/289)]

= √[(289-64)/289]

= 15/17

Thus, the value of cos b = 15/17

We know that cos (a- b) = cos a cos b + sin a sin b

Now, substitute the values for cos a, cos b, sin a and sin b in the formula, we get:

cos (a – b) = (4/5)x (15/17) + (3/5)x(8/17)

cos (a – b) = (60 + 24)/(17x 5)

cos (a – b) = 84/85

(a – b) = cos^{-1} (84/85)

Substituting the values of a and b sin^{-1} (3/5)- sin^{-1} (8/7) = cos^{-1} (84/85)

Hence proved.

**Q. 5: **Find the value of cos^{-1} (1/2) + 2 sin^{-1} (1/2).

**Solution:**

**First, solve for cos ^{-1} (1/2):**

Let us take, y = cos^{-1} (1/2)

This can be written as:

cos y = (1/2)

cos y = cos (π /3).

Thus, the range of principal value of cos^{-1} is [0, π ]

Therefore, the principal value of cos^{-1} (1/2) is π/3.

**Now, solve for sin ^{-1} (1/2):**

Let y = sin^{-1} (1/2)

sin y = 1/2

sin y = sin ( π/6)

Thus, the range of principal value of sin^{-1} is [(-π)/2, π/2 ]

Hence, the principal value of sin^{-1} (1/2) is π/6.

Now we have cos^{-1} (1/2) = π/3 & sin^{-1} (1/2) = π/6

Now, substitute the obtained values in the given formula, we get:

= cos^{-1} (1/2) + 2sin^{-1} (1/2)

= π /3 + 2( π/6)

= π/3 + π/3

= ( π+π )/3

= 2π /3

Thus, the value of cos^{-1} (1/2) + 2 sin^{-1} (1/2) is 2π /3.

### Practice Problems for Class 12 Maths Chapter 2

Solve the problems given below:

- Find the principal value of tan
^{-1}(1). - If tan
^{-1}(x − 1)/(x − 2) + tan^{-1}(x + 1)/(x + 2) = 𝜋/4, then find the value of x. - Solve: tan
^{-1}2x + tan^{-1}3x = π/4

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