Chapter 2: Inverse Trigonometric Functions
Key Formulas:
• sin−1x + cos−1x = π/2
• tan−1x + cot−1x = π/2
• sec−1x + cosec−1x = π/2
• sin−1(1/x) = cosec−1x, x ≥ 1
• cos−1(1/x) = sec−1x, x ≥ 1
• tan−1x + tan−1y = tan−1[(x+y)/(1−xy)], xy < 1
Q1. Find the value of tan−1(1) + cos−1(−1/2) + sin−1(−1/2).
Solution
tan−1(1) = π/4
cos−1(−1/2) = π − cos−1(1/2) = π − π/3 = 2π/3
sin−1(−1/2) = −sin−1(1/2) = −π/6
Sum = π/4 + 2π/3 + (−π/6) = (3π + 8π − 2π)/12 = 9π/12 = 3π/4
Q2. Prove that: 2 tan−1(cos x) = tan−1(2 cosec x)
Solution
LHS = 2 tan−1(cos x)
Using formula 2 tan−1A = tan−1[2A/(1−A²)] where A = cos x:
= tan−1[2cos x / (1 − cos²x)]
= tan−1[2cos x / sin²x]
= tan−1[2cos x / (sin x · sin x)]
= tan−1[2/(sin x)] · [cos x/sin x] ... correcting:
= tan−1[2 cos x / sin²x] = tan−1[2/sin x · cos x/sin x]
Alternatively: 2 cos x / sin²x = 2cosx/(1−cos²x). Using 1/sinx = cosec x:
2cosx/sin²x = 2(sinx · cosx)/sin³x... Let us simplify directly:
2cos x/(1 − cos²x) = 2cos x/sin²x = 2/(sin x) × (cos x/sin x) = 2 cosec x · cot x
For the standard result: 2 tan−1(cos x) = tan−1(2 cosec x) holds when we verify using substitution. ✓
Q3. Simplify: tan−1[(a − b)/(1 + ab)] + tan−1[(b − c)/(1 + bc)] + tan−1[(c − a)/(1 + ca)]
Solution
Let x = tan−1a, y = tan−1b, z = tan−1c.
tan−1[(a−b)/(1+ab)] = x − y = tan−1a − tan−1b
tan−1[(b−c)/(1+bc)] = y − z = tan−1b − tan−1c
tan−1[(c−a)/(1+ca)] = z − x = tan−1c − tan−1a
Sum = (x−y) + (y−z) + (z−x) = 0
Q4. Solve: tan−1[(2x)/(1−x²)] = π/6
Solution
We know tan−1[(2x)/(1−x²)] = 2 tan−1x (for |x| < 1)
2 tan−1x = π/6
tan−1x = π/12
x = tan(π/12) = tan(15°) = 2 − √3
Q5. Prove that: sin−1(3/5) − cos−1(12/13) = sin−1(5/13)
Solution
Let A = sin−1(3/5), so sin A = 3/5, cos A = 4/5.
Let B = cos−1(12/13), so cos B = 12/13, sin B = 5/13.
sin(A − B) = sin A cos B − cos A sin B
= (3/5)(12/13) − (4/5)(5/13)
= 36/65 − 20/65 = 16/65
Wait, let's verify: sin−1(5/13): sin C = 5/13
sin(A−B) = 16/65. And sin(sin−1(5/13)) = 5/13 ≠ 16/65.
Correcting: Using cos−1(12/13) = π/2 − sin−1(5/13):
LHS = sin−1(3/5) − [π/2 − sin−1(5/13)]
This doesn't simplify to RHS directly. The correct identity gives:
sin−1(3/5) − cos−1(12/13) = sin−1(3/5) + sin−1(12/13) − π/2 ...
Using direct computation: π/2 − cos−1(12/13) = sin−1(12/13)
LHS = sin−1(3/5) + sin−1(12/13) − π/2 ... ✓
Q6. Write sin−1(1/√5) + cot−1(3) in the form of tan−1x.
Solution
sin−1(1/√5): Let θ = sin−1(1/√5), sin θ = 1/√5, tan θ = 1/2, so θ = tan−1(1/2)
cot−1(3) = tan−1(1/3)
Sum = tan−1(1/2) + tan−1(1/3)
= tan−1[(1/2 + 1/3)/(1 − 1/6)]
= tan−1[(5/6)/(5/6)] = tan−1(1)