Sample 2025-26 Class 12 Maths Q19 Solution: sin⁻¹(√3/2) + tan⁻¹1 − sec⁻¹√2 and principal branches

The question

Select the correct answer from the options (A), (B), (C) and (D) as given below.

(A) Both (A) and (R) are true and (R) is the correct explanation of (A).

(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) (A) is true but (R) is false.

(D) (A) is false but (R) is true.

Assertion (A): Value of the expression $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) + \tan^{-1}1 - \sec^{-1}(\sqrt{2})$ is $\frac{\pi}{4}$ ⁠.

Reason (R): Principal value branch of $\sin^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and that of $\sec^{-1}x$ is $[0, \pi] - \left\{\frac{\pi}{2}\right\}$ ⁠.

Exam answer

\begin{aligned} &\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) + \tan^{-1}1 \\ &\quad - \sec^{-1}(\sqrt{2}) \\ &= \frac{\pi}{3} + \frac{\pi}{4} - \frac{\pi}{4} \\ &= \frac{\pi}{3} \ne \frac{\pi}{4} \end{aligned}

Why?

Each value is the angle in the principal value branch:

$\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ ⁠, and $\frac{\pi}{3}$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠.

$\tan\frac{\pi}{4} = 1$ ⁠, and $\frac{\pi}{4}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ⁠.

$\sec\frac{\pi}{4} = \sqrt{2}$ ⁠, because $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ ⁠, and $\frac{\pi}{4}$ lies in $[0, \pi]$ ⁠.

The last two cancel, leaving $\frac{\pi}{3}$ ⁠.

So Assertion (A) is false.

Why?

Watch out: $\frac{2\pi}{3}$ also has sine $\frac{\sqrt{3}}{2}$ ⁠, but it lies outside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠. So it is not the principal value.

Reason (R) is true: these are the principal value branches of $\sin^{-1}x$ and $\sec^{-1}x$ ⁠.

Why?

These are the standard ranges. $\frac{\pi}{2}$ is left out of the $\sec^{-1}x$ branch because $\sec\frac{\pi}{2}$ is not defined.

A is false and R is true, which is code (D).

Answer: (D) Assertion (A) is false but Reason (R) is true.

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Principal values of inverse trig functions.

Question type: Adding principal values

How to spot it: click to see

Type 4. Adding principal values: one at a time, then add

Spot it: two or three inverses joined by plus or minus signs, and the question asks for the value.

Here: Three inverses joined by plus and minus: find each one, using $\sec^{-1}\sqrt2 = \cos^{-1}\frac{1}{\sqrt2} = \frac{\pi}{4}$ ⁠, so the value is $\frac{\pi}{3} + \frac{\pi}{4} - \frac{\pi}{4} = \frac{\pi}{3}$ ⁠, not $\frac{\pi}{4}$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Q19: sin⁡−132+tan⁡−11−sec⁡−12\sin^{-1}\frac{\sqrt3}{2} + \tan^{-1}1 - \sec^{-1}\sqrt2sin−123​​+tan−11−sec−12​ and principal branches

Q19: $\sin^{-1}\frac{\sqrt3}{2} + \tan^{-1}1 - \sec^{-1}\sqrt2$ and principal branches