How Inverse Trigonometric Functions Appears in the Board Exam

A short but reliable chapter in the CBSE paper:

The marking scheme rewards the branch check: every time you cancel $f^{-1}(f(\cdot))$ or apply $f^{-1}$ to both sides, one line must confirm the angle lies in the principal range. Writing that line is the difference between full and partial credit.

Below are 12 board-style written questions with complete solutions, organised by marks, followed by a 15-question MCQ quiz.

2-Mark Questions

Question 1 — Two principal values

Find the principal values of (i) $\cos^{-1}\left(-\frac{\sqrt 3}{2}\right)$ and (ii) $\mathrm{cosec}^{-1}(-1)$.

Step 1 — (i) second-quadrant rule: $y = \pi - \cos^{-1}\frac{\sqrt 3}{2} = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \in [0, \pi]$ ✓.

Step 2 — (ii) odd-trio rule: $\mathrm{cosec}\,y = -1 \iff \sin y = -1$: $\ y = -\frac{\pi}{2}$, the endpoint of the cosecant branch ✓.

Answer: (i) $\dfrac{5\pi}{6}$, (ii) $-\dfrac{\pi}{2}$.

Question 2 — A wrap-around evaluation

Evaluate $\sin^{-1}\left(\sin\frac{5\pi}{6}\right)$.

Step 1 — branch check: $\frac{5\pi}{6} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so direct cancellation is illegal.

Step 2 — wrap: $\sin\frac{5\pi}{6} = \sin\left(\pi - \frac{5\pi}{6}\right) = \sin\frac{\pi}{6}$, and $\frac{\pi}{6}$ is inside the branch.

Answer: $\dfrac{\pi}{6}$.

Question 3 — A sum of principal values

Evaluate $\tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{\sqrt 2}\right)$.

Step 1 — each term: $\tan^{-1}(-1) = -\frac{\pi}{4}$ (odd trio); $\ \cos^{-1}\left(-\frac{1}{\sqrt 2}\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ (second quadrant).

Step 2 — add: $-\frac{\pi}{4} + \frac{3\pi}{4} = \frac{\pi}{2}$.

Answer: $\dfrac{\pi}{2}$.

Question 4 — A one-substitution simplest form

Write $\tan^{-1}\dfrac{1}{\sqrt{x^2 - 1}}$, $\ x > 1$, in simplest form.

Step 1 — substitute $x = \sec\theta$: then $\sqrt{x^2 - 1} = \tan\theta$ and the expression is $\tan^{-1}(\cot\theta) = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - \theta\right)\right) = \frac{\pi}{2} - \theta$.

Step 2 — back-substitute: $\frac{\pi}{2} - \sec^{-1}x$, which equals $\mathrm{cosec}^{-1}x$ for $x > 1$.

Answer: $\mathrm{cosec}^{-1} x$ (check at $x = 2$: both sides give $\frac{\pi}{6}$ ✓).

3-Mark Questions

Question 5 — The sine triple-angle proof

Prove that $3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$ for $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$.

Step 1 — substitute $x = \sin\theta$, so $\theta = \sin^{-1}x \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$.

Step 2 — apply the triple-angle identity: $3x - 4x^3 = 3\sin\theta - 4\sin^3\theta = \sin 3\theta$.

Step 3 — cancel legally: $3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1}x$. $\blacksquare$

Answer: proved — the restriction on $x$ exists precisely to keep $3\theta$ inside the principal branch.

Question 6 — The cosine triple-angle proof

Prove that $3\cos^{-1}x = \cos^{-1}(4x^3 - 3x)$ for $x \in \left[\frac{1}{2}, 1\right]$.

Step 1 — substitute $x = \cos\theta$, so $\theta = \cos^{-1}x \in \left[0, \frac{\pi}{3}\right]$.

Step 2 — triple-angle identity: $4x^3 - 3x = 4\cos^3\theta - 3\cos\theta = \cos 3\theta$.

Step 3 — cancel legally: $3\theta \in [0, \pi]$, the $\cos^{-1}$ range, so $\cos^{-1}(\cos 3\theta) = 3\theta = 3\cos^{-1}x$. $\blacksquare$

Question 7 — A far wrap-around

Evaluate $\tan^{-1}\left(\tan\frac{9\pi}{8}\right)$.

Step 1 — branch check: $\frac{9\pi}{8} > \frac{\pi}{2}$ — outside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Step 2 — subtract the period $\pi$: $\tan\frac{9\pi}{8} = \tan\left(\frac{9\pi}{8} - \pi\right) = \tan\frac{\pi}{8}$, and $\frac{\pi}{8}$ is inside.

Answer: $\dfrac{\pi}{8}$.

Question 8 — The other double-angle branch

Prove that $\sin^{-1}\left(2x\sqrt{1 - x^2}\right) = 2\cos^{-1}x$ for $\frac{1}{\sqrt 2} \leq x \leq 1$.

Step 1 — substitute $x = \cos\theta$, so $\theta = \cos^{-1}x \in \left[0, \frac{\pi}{4}\right]$.

Step 2 — reduce: $2x\sqrt{1 - x^2} = 2\cos\theta\sin\theta = \sin 2\theta$.

Step 3 — cancel legally: $2\theta \in \left[0, \frac{\pi}{2}\right] \subset \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\sin^{-1}(\sin 2\theta) = 2\theta = 2\cos^{-1}x$. $\blacksquare$

Answer: proved — on this window the cosine substitution is the one that keeps the doubled angle in the branch (compare the $2\sin^{-1}x$ result on $\left[-\frac{1}{\sqrt 2}, \frac{1}{\sqrt 2}\right]$).

5-Mark Questions

Question 9 — Complementary cosines

Prove that $\cos^{-1}\dfrac{3}{5} + \cos^{-1}\dfrac{4}{5} = \dfrac{\pi}{2}$.

Step 1 — name the angles: $A = \cos^{-1}\frac{3}{5}$ and $B = \cos^{-1}\frac{4}{5}$, both in $\left(0, \frac{\pi}{2}\right)$.

Step 2 — complete the 3-4-5 triangles: $\sin A = \frac{4}{5}$ and $\sin B = \frac{3}{5}$.

Step 3 — key observation: $\sin A = \cos B$, i.e. $\cos\left(\frac{\pi}{2} - A\right) = \cos B$ with both $\frac{\pi}{2} - A$ and $B$ in $\left(0, \frac{\pi}{2}\right)$; since cosine is one-one there, $B = \frac{\pi}{2} - A$. $\blacksquare$

Answer: the two angles of one 3-4-5 right triangle — of course they add to $\frac{\pi}{2}$.

Question 10 — A three-term chain

Prove that $\sin^{-1}\dfrac{3}{5} + \sin^{-1}\dfrac{8}{17} + \sin^{-1}\dfrac{36}{85} = \dfrac{\pi}{2}$.

Step 1 — combine the first two by the tangent route: $\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{3}{4}$ and $\sin^{-1}\frac{8}{17} = \tan^{-1}\frac{8}{15}$, and $$\tan(A + B) = \frac{\frac{3}{4} + \frac{8}{15}}{1 - \frac{3}{4}\cdot\frac{8}{15}} = \frac{\frac{45 + 32}{60}}{\frac{60 - 24}{60}} = \frac{77}{36} \implies A + B = \tan^{-1}\frac{77}{36}$$

Step 2 — identify the third term: the triple $36, 77, 85$ is Pythagorean ($36^2 + 77^2 = 7225 = 85^2$), so $\sin^{-1}\frac{36}{85} = \tan^{-1}\frac{36}{77}$.

Step 3 — the two tangents are reciprocals: $\tan^{-1}\frac{77}{36} + \tan^{-1}\frac{36}{77} = \frac{\pi}{2}$ (complementary angles in a right triangle with legs 77 and 36). $\blacksquare$

Question 11 — An equation in board format

Solve $2\tan^{-1}(\cos x) = \tan^{-1}(2\,\mathrm{cosec}\,x)$.

Step 1 — double formula on the left: $$2\tan^{-1}(\cos x) = \tan^{-1}\frac{2\cos x}{1 - \cos^2 x} = \tan^{-1}\frac{2\cos x}{\sin^2 x}$$

Step 2 — equate the arguments ($\tan^{-1}$ is one-one): $$\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x} \implies 2\cos x \sin x = 2\sin^2 x \implies \sin x(\cos x - \sin x) = 0$$

Step 3 — reject $\sin x = 0$ (it makes $\mathrm{cosec}\,x$ undefined), leaving $\tan x = 1$.

Answer: $x = \dfrac{\pi}{4}$ — with the rejection step written out for the method marks.

Question 12 — The nested-roots reduction

Prove that $\cot^{-1}\left(\dfrac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right) = \dfrac{x}{2}$ for $x \in \left(0, \frac{\pi}{4}\right)$.

Step 1 — perfect squares under the roots: $$1 \pm \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} \pm 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} \pm \sin\frac{x}{2}\right)^2$$

Step 2 — drop the moduli (on $0 < x < \frac{\pi}{4}$, both $\cos\frac{x}{2} + \sin\frac{x}{2}$ and $\cos\frac{x}{2} - \sin\frac{x}{2}$ are positive): $$\frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) + \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) - \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)} = \frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}$$

Step 3 — cancel legally: $\frac{x}{2} \in \left(0, \frac{\pi}{8}\right) \subset (0, \pi)$, so $\cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}$. $\blacksquare$

Presentation tip: in every solution above, one line justifies dropping a modulus or cancelling an inverse — those single lines carry the reasoning marks in this chapter.