# NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 2: Inverse Trigonometric Functions (Part I)

NCERT Solutions for CBSE 12th Mathematics, Chapter 2: Inverse Trigonometric Functions are available here. In this article, you will get solutions of exercise 2.1 (from question number 1 to question number 7). Most of the questions given in this exercise are based on basic concepts related to inverse trigonometric functions. These questions are important CBSE Class 12 Maths board exam 2018 and other competitive exams like JEE Main, WBJEE etc.

NCERT Solutions for CBSE Class 12th Maths, Chapter 2: Inverse Trigonometric Functions (Exercise 2.1) are given below

**Question 1:** sin^{‒1 }(‒1/2)

**Solution 1:**

Let sin^{‒1 }(‒1/2) = *y*

⇒ sin *y* = ‒1/2

⇒ sin *y* = ‒sin (π/6)

⇒* y* = π/6

**NCERT Exemplar Class 12 Maths – Chapter 2**

**Question 2:** cos^{‒1}(√3/2)

**Solution 2:**

cos^{‒1}(√3/2) = *y*

⇒ cos *y* = √3/2

⇒ cos *y* =cos (π/6)

⇒ *y* =cos (π/6)

**Question 3:** cosec^{‒1 }(2)

**Solution 3:**

Let *y* = cosec^{‒1} (2)

⇒ Cosec *y* = 2

⇒ Cosec *y* = cosec (π/6)

⇒ *y* = π/6

**Question 4:** tan^{‒1} (‒√3)

**Solution 4:**

Let tan^{‒1} (‒√3) = *y*

⇒ tan y = ‒ tan (π/3)

⇒ tan *y* = tan (‒π/3)

⇒ *y* = ‒ π/3

**Question 5:** cos^{‒1} (‒1/2)

**Solution 5:**

Let cos^{‒1} (‒1/2) = *y*

⇒cos *y* = ‒ (1/2)

⇒cos *y* = ‒ cos (π/3)

⇒cos *y* = ‒ cos (π ‒ π/3)

⇒cos *y* = ‒ cos (2π/3)

⇒ *y* = (2π/3)

**Question 6:** tan^{‒1} (‒1)

**Solution 6:**

Let tan^{‒1} (‒1) = *y*

⇒ tan *y* = ‒1

⇒ tan *y* = ‒ tan (π/4)

⇒ tan *y* = tan (‒π/4)

⇒ *y* = ‒π/4

**Question 7:** sec^{‒1} (2/√3)

**Solution 7:**

Let sec^{‒1} (2/√3) = *y*

⇒ sec *y* = 2/√3

⇒ sec *y* = sec(π/6)

⇒ y = (π/6).