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

NCERT Solutions for CBSE 12th Mathematics, Chapter 2: Inverse Trigonometric Functions are available here. Here, you will get solutions of exercise 2.1 (from question number 8 to question number 14). 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 UPSEE, JEE Main, WBJEE etc.

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

**Question 8**: cot^{‒1} (√3)

**Solution 8:**

Let cot^{‒1}(√3) = *y*

⇒cot *y* = √3

⇒cot *y* = cot (π/6)

⇒*y* = π/6

**Question 9**: cos^{‒1}(‒1/√2)

**Solution 9:**

cos^{‒1} (‒1/√2) = y

⇒ cos *y* = ‒1/√2

⇒ cos y = ‒ cos (π/4)

⇒ cos y = cos (π ‒ π/4)

⇒ cos y = cos (3π/4)

⇒ y = 3π/4

**Question** **10:** cosec^{‒1}(‒√2)

**Solution 10:**

Let cosec^{‒1} (‒√2) = *y*

⇒ cosec *y* = ‒√2

⇒ cosec *y* = cosec (‒π/4)

⇒* y* = ‒π/4

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

**Question 11:** Find the value of tan^{‒1} (1) + cos^{‒1} (‒1/2) + sin^{‒1} (‒1/2)

**Solution 11:**

Let tan^{‒1} (1) = *A*

⇒ tan *A* = 1

⇒ tan *A* = tan (π/4)

⇒ *A* = π/4

Let cos^{‒1} (‒1/2) = *B*

⇒ cos *B* = (‒1/2)

⇒ cos *B* = ‒ cos (π/3)

⇒ cos *B* = cos (π ‒ π/3)

⇒ cos *B* = cos (2π/3)

⇒ *B* = 2π/3

Let sin^{‒1} (‒1/2) = *C*

⇒ sin *c* = (‒1/2)

⇒ sin *c* = ‒ sin (π/6)

⇒ sin *c* = sin (‒π/6)

⇒ *c* = ‒π/6

tan^{‒1} (1) + cos^{‒1} (‒1/2) + sin^{‒1} (‒1/2) = (π/4) + (2π/3) ‒ (π/6) = (3π + 8π ‒ 2 π)/12 = 3π/12.

**Question 12: **Find the value of** **

cos** ^{‒1}**(1/2) + 2sin

^{‒1}(1/2)

**Solution 12:**

Let cos** ^{‒1}**(1/2) =

*A*

⇒cos A = ½

⇒cos A = cos (π/3)

⇒ A = π/3

Let sin** ^{‒1}**(1/2) =

*B*

⇒sin B = ½

⇒sin B = sin (π/6)

⇒ B = (π/6)

cos** ^{‒1}**(1/2) + 2sin

^{‒1}(1/2) = π/3 + 2(π/6) = π/3 + 2 (π/3) = 2 π/3

**Question 13:** Find the value of if sin^{−1} *x *= *y*, then

(A) 0 ≤ *y* ≤ π

(B) ‒ π/2 ≤ *y *≤ π/2

(C) 0 < y < π

(D) ‒ π/2 < *y* < π/2

**Solution 13:**

sin^{‒1} (*x*) = *y*

Range of the principal value branch of sin^{−1} is [‒ π/2, π/2]

Therefore, ‒ π/2 < *y* < π/2 i.e., option (D) is correct

**Question 14**. tan ^{‒1}√3 ‒ sec ^{‒1} (‒2) is equal to

(A) π

(B) ‒ π/3

(C) π/3

(D) 2π/3

**Solution 14:**

Let tan^{‒1 }√3 = A

⇒ tan *A* = √3

⇒ tan *A* = tan π/3

⇒ *A* = π/3

Let sec^{‒1} (‒2) = B

⇒ sec *B* = ‒2

⇒ sec *B *= ‒ sec (π/3)

⇒ sec *B* = sec (π ‒ π/3)

⇒ sec *B* = sec (2π/3)

⇒ *B *= (2π/3)

Now, tan^{‒1 }√3 ‒ sec^{‒1}(‒2) = (π/3) ‒ (2π/3) = ‒ π/3.