Ch 2: Inverse Trigonometric Functions
Master Principal Value Branches, Simplification, and the complete Miscellaneous Exercise. Highly optimized for CBSE Board Exams.
Exercise 2.1 Solutions
⚠️ Board Tip: Memorize the domain and range (Principal Value Branch) of all 6 inverse trig functions. 1 mark MCQs strictly rely on this.
Find the principal values of the following (Q1 to Q10):
Q1. sin⁻¹(-1/2)
Solution:
Let y = sin⁻¹(-1/2). Then, sin(y) = -1/2.
We know that sin(π/6) = 1/2. So, sin(-π/6) = -1/2.
The principal value branch of sin⁻¹ is [-π/2, π/2].
Since -π/6 ∈ [-π/2, π/2], the principal value is -π/6.
Q2. cos⁻¹(√3/2)
Solution:
Let y = cos⁻¹(√3/2). Then, cos(y) = √3/2.
We know that cos(π/6) = √3/2.
The principal value branch of cos⁻¹ is [0, π].
Since π/6 ∈ [0, π], the principal value is π/6.
Q5. cos⁻¹(-1/2)
Solution:
Let y = cos⁻¹(-1/2). Then, cos(y) = -1/2.
cos(y) = -cos(π/3) = cos(π - π/3) = cos(2π/3).
The principal value branch of cos⁻¹ is [0, π].
Since 2π/3 ∈ [0, π], the principal value is 2π/3.
Q11. Find the value of: tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/2)
Solution:
1. tan⁻¹(1) = π/4 (Since tan(π/4) = 1 and π/4 ∈ (-π/2, π/2))
2. cos⁻¹(-1/2) = 2π/3 (Calculated in Q5)
3. sin⁻¹(-1/2) = -π/6 (Calculated in Q1)
Sum = π/4 + 2π/3 - π/6
LCM of 4, 3, 6 is 12.
Sum = (3π + 8π - 2π) / 12 = 9π / 12 = 3π/4.
Q13. If sin⁻¹x = y, then:
Solution:
The range of the principal value branch of sin⁻¹x is [-π/2, π/2].
Therefore, y must lie within this interval.
Correct Option: (B) -π/2 ≤ y ≤ π/2
Exercise 2.2 Solutions
Q1. Prove that: 3sin⁻¹x = sin⁻¹(3x - 4x³), x ∈ [-1/2, 1/2]
Solution:
Let x = sin(θ). Therefore, θ = sin⁻¹x.
RHS = sin⁻¹(3x - 4x³)
= sin⁻¹(3sin(θ) - 4sin³(θ))
= sin⁻¹(sin(3θ)) [Using formula sin(3θ) = 3sinθ - 4sin³θ]
= 3θ
= 3sin⁻¹x = LHS. Hence Proved.
Q3. Prove that: tan⁻¹(2/11) + tan⁻¹(7/24) = tan⁻¹(1/2)
Solution:
Using the formula: tan⁻¹A + tan⁻¹B = tan⁻¹[(A + B) / (1 - AB)]
LHS = tan⁻¹[ (2/11 + 7/24) / (1 - (2/11)(7/24)) ]
= tan⁻¹[ ((48 + 77) / 264) / (1 - 14/264) ]
= tan⁻¹[ (125 / 264) / (250 / 264) ]
= tan⁻¹(125 / 250)
= tan⁻¹(1/2) = RHS. Hence Proved.
Next Chapters
Ch 3: Matrices Ch 4: DeterminantsQ5. Write the function in the simplest form: tan⁻¹( (√(1+x²) - 1) / x ), x ≠ 0
Solution:
Put x = tan(θ). Then θ = tan⁻¹x.
Expression = tan⁻¹[ (√(1+tan²θ) - 1) / tanθ ]
= tan⁻¹[ (secθ - 1) / tanθ ] [Since 1+tan²θ = sec²θ]
= tan⁻¹[ (1/cosθ - 1) / (sinθ/cosθ) ]
= tan⁻¹[ (1 - cosθ) / sinθ ]
Use half-angle formulas: 1 - cosθ = 2sin²(θ/2) and sinθ = 2sin(θ/2)cos(θ/2)
= tan⁻¹[ (2sin²(θ/2)) / (2sin(θ/2)cos(θ/2)) ]
= tan⁻¹[ sin(θ/2) / cos(θ/2) ]
= tan⁻¹[ tan(θ/2) ]
= θ/2
Substitute back θ = tan⁻¹x. The simplest form is: (1/2) tan⁻¹x
Q13. Find the value of: tan[ (1/2) [ sin⁻¹(2x / 1+x²) + cos⁻¹(1-y² / 1+y²) ] ]
Solution:
We know standard identities:
sin⁻¹(2x / 1+x²) = 2tan⁻¹x
cos⁻¹(1-y² / 1+y²) = 2tan⁻¹y
Substitute these into the expression:
= tan[ (1/2) (2tan⁻¹x + 2tan⁻¹y) ]
= tan[ tan⁻¹x + tan⁻¹y ]
= tan[ tan⁻¹( (x + y) / (1 - xy) ) ]
= (x + y) / (1 - xy)
Miscellaneous Exercise (All Questions)
Q1. Find the value of cos⁻¹(cos(13π/6))
Solution:
We know cos⁻¹(cos θ) = θ only if θ ∈ [0, π].
13π/6 = 2π + π/6, which is outside [0, π].
cos(13π/6) = cos(2π + π/6) = cos(π/6)
Therefore, cos⁻¹(cos(13π/6)) = cos⁻¹(cos(π/6)) = π/6.
Q2. Find the value of tan⁻¹(tan(7π/6))
Solution:
We know tan⁻¹(tan θ) = θ only if θ ∈ (-π/2, π/2).
7π/6 = π + π/6, which is outside (-π/2, π/2).
tan(7π/6) = tan(π + π/6) = tan(π/6) (Since tan is positive in 3rd quadrant)
Therefore, tan⁻¹(tan(7π/6)) = tan⁻¹(tan(π/6)) = π/6.
Q3. Prove that: 2sin⁻¹(3/5) = tan⁻¹(24/7)
Solution:
Let sin⁻¹(3/5) = x. Then sin(x) = 3/5.
Using a right triangle, Perpendicular = 3, Hypotenuse = 5. Base = √(5² - 3²) = 4.
So, tan(x) = P/B = 3/4. Therefore, x = tan⁻¹(3/4).
Now, LHS = 2tan⁻¹(3/4).
Using formula 2tan⁻¹A = tan⁻¹(2A / (1 - A²)):
= tan⁻¹[ (2(3/4)) / (1 - (3/4)²) ]
= tan⁻¹[ (3/2) / (1 - 9/16) ]
= tan⁻¹[ (3/2) / (7/16) ]
= tan⁻¹(24/7) = RHS. Proved.
Q4. Prove that: sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36)
Solution:
Convert both to tan⁻¹:
Let x = sin⁻¹(8/17) → sin(x) = 8/17. Base = √(17²-8²) = 15. So, tan(x) = 8/15.
Let y = sin⁻¹(3/5) → sin(y) = 3/5. Base = √(5²-3²) = 4. So, tan(y) = 3/4.
LHS = tan⁻¹(8/15) + tan⁻¹(3/4)
= tan⁻¹[ (8/15 + 3/4) / (1 - (8/15)(3/4)) ]
= tan⁻¹[ (32 + 45)/60 / (1 - 24/60) ]
= tan⁻¹[ (77/60) / (36/60) ]
= tan⁻¹(77/36) = RHS. Proved.
Q8. Prove that: tan⁻¹(1/5) + tan⁻¹(1/7) + tan⁻¹(1/3) + tan⁻¹(1/8) = π/4
Solution:
Group the terms into two pairs: [tan⁻¹(1/5) + tan⁻¹(1/7)] + [tan⁻¹(1/3) + tan⁻¹(1/8)]
First pair:
= tan⁻¹[ (1/5 + 1/7) / (1 - 1/35) ] = tan⁻¹[ (12/35) / (34/35) ] = tan⁻¹(12/34) = tan⁻¹(6/17)
Second pair:
= tan⁻¹[ (1/3 + 1/8) / (1 - 1/24) ] = tan⁻¹[ (11/24) / (23/24) ] = tan⁻¹(11/23)
Now add the results: tan⁻¹(6/17) + tan⁻¹(11/23)
= tan⁻¹[ (6/17 + 11/23) / (1 - (6/17)(11/23)) ]
= tan⁻¹[ (138 + 187)/391 / (1 - 66/391) ]
= tan⁻¹[ (325/391) / (325/391) ]
= tan⁻¹(1) = π/4 = RHS. Proved.
Q9. Prove that: tan⁻¹(√x) = (1/2)cos⁻¹( (1-x)/(1+x) )
Solution:
Let RHS = (1/2)cos⁻¹( (1-x)/(1+x) ). Put x = tan²(θ), so √x = tan(θ) and θ = tan⁻¹(√x).
RHS = (1/2)cos⁻¹( (1 - tan²θ) / (1 + tan²θ) )
We know the identity: cos(2θ) = (1 - tan²θ) / (1 + tan²θ)
RHS = (1/2)cos⁻¹(cos(2θ))
= (1/2)(2θ)
= θ
= tan⁻¹(√x) = LHS. Proved.
Q10. Prove that: cot⁻¹[ (√(1+sin x) + √(1-sin x)) / (√(1+sin x) - √(1-sin x)) ] = x/2, x ∈ (0, π/4)
Solution:
We use the identity: 1 ± sin(x) = (cos(x/2) ± sin(x/2))².
√(1+sin x) = cos(x/2) + sin(x/2)
√(1-sin x) = cos(x/2) - sin(x/2)
Substitute in the numerator:
Num = (cos(x/2) + sin(x/2)) + (cos(x/2) - sin(x/2)) = 2cos(x/2)
Substitute in the denominator:
Den = (cos(x/2) + sin(x/2)) - (cos(x/2) - sin(x/2)) = 2sin(x/2)
LHS = cot⁻¹[ 2cos(x/2) / 2sin(x/2) ]
= cot⁻¹[ cot(x/2) ]
= x/2 = RHS. Proved.
Q11. Prove that: tan⁻¹[ (√(1+x) - √(1-x)) / (√(1+x) + √(1-x)) ] = π/4 - (1/2)cos⁻¹x
Solution:
Put x = cos(2θ). Then 2θ = cos⁻¹x → θ = (1/2)cos⁻¹x.
1 + x = 1 + cos(2θ) = 2cos²θ → √(1+x) = √2 cosθ
1 - x = 1 - cos(2θ) = 2sin²θ → √(1-x) = √2 sinθ
Substitute into LHS:
= tan⁻¹[ (√2cosθ - √2sinθ) / (√2cosθ + √2sinθ) ]
Cancel √2 and divide numerator and denominator by cosθ:
= tan⁻¹[ (1 - tanθ) / (1 + tanθ) ]
This is the expansion of tan(π/4 - θ):
= tan⁻¹[ tan(π/4 - θ) ]
= π/4 - θ
= π/4 - (1/2)cos⁻¹x = RHS. Proved.
Q12. Prove that: 9π/8 - 9/4 sin⁻¹(1/3) = 9/4 sin⁻¹(2√2/3)
Solution:
LHS = 9/4 [ π/2 - sin⁻¹(1/3) ]
We know that sin⁻¹x + cos⁻¹x = π/2. Therefore, π/2 - sin⁻¹x = cos⁻¹x.
LHS = 9/4 cos⁻¹(1/3)
Now, convert cos⁻¹(1/3) into sin⁻¹:
Let cos⁻¹(1/3) = θ → cosθ = 1/3. (Base=1, Hypotenuse=3)
Perpendicular = √(3² - 1²) = √8 = 2√2.
So, sinθ = 2√2 / 3 → θ = sin⁻¹(2√2/3).
Substitute back: LHS = 9/4 sin⁻¹(2√2/3) = RHS. Proved.
Q13 & Q14. Solve for x (Equations):
Q13. 2tan⁻¹(cosx) = tan⁻¹(2cosecx)
Using 2tan⁻¹A = tan⁻¹(2A / 1-A²):
tan⁻¹(2cosx / (1-cos²x)) = tan⁻¹(2/sinx)
2cosx / sin²x = 2/sinx
Cancel 2/sinx (assuming sinx ≠ 0):
cosx / sinx = 1 → cotx = 1 → x = π/4
Q14. tan⁻¹( (1-x)/(1+x) ) = (1/2)tan⁻¹x, (x > 0)
Put x = tanθ:
tan⁻¹[ (1-tanθ)/(1+tanθ) ] = (1/2)θ
tan⁻¹[ tan(π/4 - θ) ] = θ/2
π/4 - θ = θ/2
π/4 = 3θ/2 → θ = π/6
Since x = tanθ → x = tan(π/6) = 1/√3
Q15, 16 & 17. Multiple Choice Questions
Q15: sin(tan⁻¹x), |x| < 1 is equal to:
Let tan⁻¹x = θ → tanθ = x/1 (P/B). H = √(1+x²).
sinθ = P/H = x / √(1+x²). Ans: (D)
Q16: sin⁻¹(1-x) - 2sin⁻¹x = π/2, then x is equal to:
Put x = 0. LHS = sin⁻¹(1) - 0 = π/2 = RHS. Correct.
Put x = 1/2. LHS = sin⁻¹(1/2) - 2sin⁻¹(1/2) = -sin⁻¹(1/2) = -π/6 ≠ RHS. False.
Ans: (C) 0
Q17: tan⁻¹(x/y) - tan⁻¹((x-y)/(x+y)) is equal to:
Divide numerator and denominator of 2nd term by y: tan⁻¹[ (x/y - 1) / (x/y + 1) ]
= tan⁻¹[ (x/y) ] - tan⁻¹(1)
The expression becomes: tan⁻¹(x/y) - [ tan⁻¹(x/y) - π/4 ] = π/4.
Ans: (C) π/4