Practice Exercise 2.1

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This page provides comprehensive Class 12 - Practice Exercise 2.1. Free NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2-1. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.

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 = -Ï€/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(Ï€/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.
Ï€/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 = -Ï€/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 = 3Ï€/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 = -Ï€/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(Ï€/4) = 1.
For cot, negative values are in 2nd quadrant: π - π/4 = 3π/4.
Principal Value = 3Ï€/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 = 5Ï€/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 = -Ï€/6.
Principal Value = -Ï€/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 = 2Ï€/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 = 2Ï€/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) -Ï€/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
Ï€/4 - 3Ï€/4 = -2Ï€/4 = -Ï€/2.
Correct Option: (B)
Chapter 1 Miscellaneous Exercise 2.2