CBSE Class 12 Mathematics Inverse Trigonometric Functions Case Studies

Answer: (b) \( \sin^{-1}\left(\frac{1}{2}\right) \)

Explanation:
1. In right △ADB: BD = 30 m, AD = \( 30\sqrt{3} \) m, so AB = \( \sqrt{30^2 + (30\sqrt{3})^2} = \sqrt{3600} \) = 60 m.
2. sin α = \( \frac{BD}{AB} = \frac{30}{60} = \frac{1}{2} \)
3. α = \( \sin^{-1}\left(\frac{1}{2}\right) \) (= 30°)

Teacher's Note:
1. Options with sin⁻¹ of 2 or \( \frac{2}{\sqrt{3}} \) are impossible, since sin⁻¹ needs an input in [−1, 1].
2. Using tan α = \( \frac{30}{30\sqrt{3}} = \frac{1}{\sqrt{3}} \) first gives α = 30° quickly, then convert to any ratio.

Answer: (c) \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \)

Explanation:
1. In right △ADB, AD = \( 30\sqrt{3} \) m and AB = 60 m.
2. cos α = \( \frac{AD}{AB} = \frac{30\sqrt{3}}{60} = \frac{\sqrt{3}}{2} \)
3. α = \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \) (= 30°)

Teacher's Note:
1. The same angle can be written as sin⁻¹, cos⁻¹ or tan⁻¹ of different values; all equal 30° here.
2. Check: \( \sin^2\alpha + \cos^2\alpha = \frac{1}{4} + \frac{3}{4} = 1 \), confirming the pair of values.

Answer: (d) \( \tan^{-1}(\sqrt{3}) \)

Explanation:
1. DC = AC − AD = \( 40\sqrt{3} - 30\sqrt{3} = 10\sqrt{3} \) m
2. In right △CDB: tan β = \( \frac{BD}{DC} = \frac{30}{10\sqrt{3}} = \sqrt{3} \)
3. β = \( \tan^{-1}(\sqrt{3}) \) (= 60°)

Teacher's Note:
1. The distance 40√3 m is between the two men, so subtract AD to get the second man's distance from the temple.
2. The nearer observer sees the top at the larger angle, so β (60°) > α (30°) is a good sense check.

Answer: (c) \( \frac{\pi}{2} \)

Explanation:
1. From the previous parts, α = \( \frac{\pi}{6} \) and β = \( \frac{\pi}{3} \).
2. In △ABC: ∠ABC = π − (α + β) = \( \pi - \left(\frac{\pi}{6} + \frac{\pi}{3}\right) \)
3. ∠ABC = \( \frac{\pi}{2} \)

Teacher's Note:
1. Angle sum of a triangle is π, so the angle at the top is π minus the two angles of elevation.
2. Check: BC = 20√3 m and AB² + BC² = 3600 + 1200 = 4800 = (40√3)² = AC², so the angle at B is indeed a right angle.

Answer: (c) [−1, 1], [0, π]

Explanation:
1. cos x takes values only in [−1, 1], so cos⁻¹x is defined for x in [−1, 1] (end points included).
2. The principal branch of cos⁻¹ is [0, π], with cos⁻¹(1) = 0 and cos⁻¹(−1) = π.
3. Domain = [−1, 1], Range = [0, π].

Teacher's Note:
1. Closed brackets matter: cos⁻¹(1) = 0 and cos⁻¹(−1) = π are both attained.
2. \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) is the range of sin⁻¹x, a common mix-up with cos⁻¹x.