NCERT Solutions for Class 12 Maths Exercise 2.1: Principal Values

Overview

Work through Class 12 Maths NCERT Exercise 2.1 on principal values of inverse trigonometric functions, with step-by-step solutions for CBSE revision.

Inverse Trigonometric Functions: Principal Values.
Detailed solutions for similar practice problems.

Chapter Index Start Solving

Q1 Sin Q2 Cos Q3 Cosec Q4 Tan Q5 Cos Q6 Tan Q7 Sec Q8 Cot Q9 Cos Q10 Cosec Q11 Mixed Q12 Mixed Q13 MCQ Q14 MCQ

Q1 - Q10. Find Principal Values

Similar to Q1

Find the principal value of: sin⁻¹(-√3/2)

View Step-by-Step Solution

Step 1: Identify Range

The principal value branch of sin⁻¹ is [-π/2, π/2].

Step 2: Solve

Let sin⁻¹(-√3/2) = y.
Then sin y = -√3/2.
We know sin(π/3) = √3/2.
Since sin(-x) = -sin(x), sin(-π/3) = -√3/2.
-π/3 lies in [-π/2, π/2].

Principal Value = -π/3

Similar to Q2

Find the principal value of: cos⁻¹(1/2)

View Solution

Range

Principal branch of cos⁻¹ is [0, π].

Solution

cos y = 1/2.
We know cos(π/3) = 1/2.
π/3 lies in [0, π].

Principal Value = π/3

Similar to Q3

Find the principal value of: cosec⁻¹(√2)

View Solution

Range

[-π/2, π/2] - {0}

Solution

cosec y = √2.
Since sin(π/4) = 1/√2, cosec(π/4) = √2.
π/4 is in the valid range.

Principal Value = π/4

Similar to Q4

Find the principal value of: tan⁻¹(-1/√3)

View Solution

Range

(-π/2, π/2)

Solution

tan y = -1/√3.
tan(π/6) = 1/√3.
Since tan(-x) = -tan(x), tan(-π/6) = -1/√3.

Principal Value = -π/6

Similar to Q5

Find the principal value of: cos⁻¹(-1/√2)

View Solution

Range

[0, π]

Solution

cos y = -1/√2.
cos(π/4) = 1/√2.
Since cos is negative in 2nd quadrant: cos(π - π/4) = cos(3π/4) = -1/√2.

Principal Value = 3π/4

Similar to Q6

Find the principal value of: tan⁻¹(-√3)

View Solution

Range

(-π/2, π/2)

Solution

tan y = -√3.
tan(π/3) = √3.
tan(-π/3) = -√3.

Principal Value = -π/3

Similar to Q7

Find the principal value of: sec⁻¹(2)

View Solution

Range

[0, π] - {π/2}

Solution

sec y = 2.
cos y = 1/2.
y = π/3.

Principal Value = π/3

Similar to Q8

Find the principal value of: cot⁻¹(-1)

View Solution

Range

(0, π)

Solution

cot y = -1.
cot(π/4) = 1.
For cot, negative values are in 2nd quadrant: π - π/4 = 3π/4.

Principal Value = 3π/4

Similar to Q9

Find the principal value of: cos⁻¹(-√3/2)

View Solution

Range

[0, π]

Solution

cos y = -√3/2.
cos(π/6) = √3/2.
2nd quadrant: π - π/6 = 5π/6.

Principal Value = 5π/6

Similar to Q10

Find the principal value of: cosec⁻¹(-2)

View Solution

Range

[-π/2, π/2] - {0}

Solution

cosec y = -2.
sin y = -1/2.
y = -π/6.

Principal Value = -π/6

Q11 - Q12. Evaluate Expressions

Similar to Q11

Find value of: tan⁻¹(√3) + sec⁻¹(-2) + cosec⁻¹(-2/√3)

View Solution

Step 1: tan⁻¹(√3)

= π/3

Step 2: sec⁻¹(-2)

sec y = -2 (2nd quad for sec⁻¹). y = π - π/3 = 2π/3.

Step 3: cosec⁻¹(-2/√3)

cosec y = -2/√3. sin y = -√3/2. y = -π/3.

Step 4: Sum

π/3 + 2π/3 - π/3 = 2π/3.

Answer = 2π/3

Similar to Q12

Find value of: cos⁻¹(1/2) + 2sin⁻¹(1/2)

View Solution

Step 1: Values

cos⁻¹(1/2) = π/3.
sin⁻¹(1/2) = π/6.

Step 2: Calculate

π/3 + 2(π/6) = π/3 + π/3 = 2π/3.

Answer = 2π/3

Q13 - Q14. MCQs

Similar to Q13

If sin⁻¹x = y, then:
(A) 0 ≤ y ≤ π
(B) -π/2 ≤ y ≤ π/2
(C) 0 < y < π
(D) -π/2 < y < π/2

View Solution

Explanation

The principal value branch (range) of sin⁻¹x is the closed interval [-π/2, π/2].

Correct Option: (B)

Similar to Q14

The value of tan⁻¹(1) - cot⁻¹(-1) is equal to:
(A) π
(B) -π/2
(C) 0
(D) 2√3

View Solution

Step 1: tan⁻¹(1)

= π/4

Step 2: cot⁻¹(-1)

= π - π/4 = 3π/4 (cot⁻¹ range is (0, π)).

Step 3: Subtract

π/4 - 3π/4 = -2π/4 = -π/2.

Correct Option: (B)