Inverse Trigonometric Functions — Class 12 Maths

🎯 Problem aur Solution

Agar tumse kaha jaaye "sin(x) = 1/2" toh x kya hai? Tum bolo "x = 30°" — lekin yeh bhi possible hai x = 150°, ya 390°, ya... infinite values! Ek trigonometric function many-to-one hai — alag alag angles ka same sin value ho sakta hai. Toh f⁻¹ exist kaise karega? Solution: domain restrict karo ek specific interval par jahan function one-one ban jaaye. Isi restricted domain par inverse trig functions define hote hain!

1. Domain aur Range — Principal Value Branch

Inverse trig functions define karne ke liye hum trig functions ko restrict karte hain unke principal value branch par — ek aisa interval jahan function strictly one-one aur onto ho.

sin⁻¹(x)

Domain: [-1, 1]

Range: [-π/2, π/2]

First & Fourth quadrant

cos⁻¹(x)

Domain: [-1, 1]

Range: [0, π]

First & Second quadrant

tan⁻¹(x)

Domain: ℝ (all real)

Range: (-π/2, π/2)

Open interval

cosec⁻¹(x)

Domain: |x| ≥ 1

Range: [-π/2,π/2] \ {0}

0 exclude

sec⁻¹(x)

Domain: |x| ≥ 1

Range: [0,π] \ {π/2}

π/2 exclude

cot⁻¹(x)

Domain: ℝ

Range: (0, π)

Open interval

2. Important Properties aur Identities

/* Complementary angle identities */ sin⁻¹x + cos⁻¹x = π/2 (for x ∈ [-1,1]) tan⁻¹x + cot⁻¹x = π/2 (for x ∈ ℝ) sec⁻¹x + cosec⁻¹x = π/2 (for |x| ≥ 1) /* Negative argument */ sin⁻¹(-x) = -sin⁻¹x cos⁻¹(-x) = π - cos⁻¹x tan⁻¹(-x) = -tan⁻¹x /* Double angle */ 2sin⁻¹x = sin⁻¹(2x√(1-x²)) 2cos⁻¹x = cos⁻¹(2x²-1) 2tan⁻¹x = tan⁻¹(2x/(1-x²)) /* Addition formulae */ tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1-xy)) [xy < 1]

3. Principal Values — Worked Examples

✅ Example 1: sin⁻¹(1/2)

Humein angle θ chahiye jahan sin θ = 1/2 aur θ ∈ [-π/2, π/2]

sin(π/6) = 1/2 aur π/6 ∈ [-π/2, π/2] ✓

∴ sin⁻¹(1/2) = π/6 = 30°

✅ Example 2: cos⁻¹(-1/2)

Humein θ chahiye jahan cos θ = -1/2 aur θ ∈ [0, π]

cos(2π/3) = -1/2 aur 2π/3 ∈ [0, π] ✓

∴ cos⁻¹(-1/2) = 2π/3 = 120°

✅ Example 3: tan⁻¹(√3)

tan(π/3) = √3 aur π/3 ∈ (-π/2, π/2) ✓

∴ tan⁻¹(√3) = π/3 = 60°

4. Important Standard Values Table

Value	sin⁻¹	cos⁻¹	tan⁻¹
0	0	π/2	0
1/2	π/6	π/3	π/6 (approx)
1/√2	π/4	π/4	π/4
√3/2	π/3	π/6	—
1	π/2	0	π/4 (approx)
-1	-π/2	π	—
√3	—	—	π/3

5. Practice Questions

Q1. Prove karo: sin⁻¹(3/5) + sin⁻¹(4/5) = π/2

Hint dekhein

Let α = sin⁻¹(3/5), β = sin⁻¹(4/5). Then sinα = 3/5, cosα = 4/5 aur sinβ = 4/5, cosβ = 3/5. sin(α+β) = sinα.cosβ + cosα.sinβ = (3/5)(3/5) + (4/5)(4/5) = 9/25 + 16/25 = 1. So α+β = π/2 ✓

Q2. Find the value of: tan⁻¹(1) + tan⁻¹(√3) + tan⁻¹(1/√3)

Solution dekhein

= π/4 + π/3 + π/6 = 3π/12 + 4π/12 + 2π/12 = 9π/12 = 3π/4

Q3. Simplify: sin⁻¹(sin(3π/4))