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CBSE Sample Paper 2025-26 · Section B · 2 marks

Q21: Evaluate tan⁡(tan⁡−1(−1)+π3)\tan\left(\tan^{-1}(-1) + \frac{\pi}{3}\right)⁠, or the domain of cos⁡−1(3x−2)\cos^{-1}(3x - 2)

Sample 2025-26Ch 2 · Inverse Trigonometric Functions
The question

(a) Evaluate tan⁡(tan⁡−1(−1)+π3)\tan\left(\tan^{-1}(-1) + \dfrac{\pi}{3}\right)

OR

(b) Find the domain of cos⁡−1(3x−2)\cos^{-1}(3x - 2)

Exam answer

(a)

The principal value of tan⁡−1(−1)\tan^{-1}(-1) is −π4-\frac{\pi}{4}⁠.

Why?

The principal value of tan⁡−1\tan^{-1} is the one angle in this interval with the given tangent:

tan⁡−1x∈(−π2,π2)\tan^{-1}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

tan⁡π4=1\tan\frac{\pi}{4} = 1 and tangent is an odd function, so tan⁡(−π4)=−1\tan\left(-\frac{\pi}{4}\right) = -1⁠, and −π4-\frac{\pi}{4} lies in the interval.

Watch out: 3π4\frac{3\pi}{4} also has tangent −1-1⁠, but it is outside the principal interval.

tan⁡(tan⁡−1(−1)+π3)=tan⁡(−π4+π3)=tan⁡π3−tan⁡π41+tan⁡π3tan⁡π4=3−11+3\begin{aligned} &\tan\left(\tan^{-1}(-1) + \frac{\pi}{3}\right) \\ &= \tan\left(-\frac{\pi}{4} + \frac{\pi}{3}\right) \\ &= \frac{\tan\frac{\pi}{3} - \tan\frac{\pi}{4}}{1 + \tan\frac{\pi}{3}\tan\frac{\pi}{4}} \\ &= \frac{\sqrt3 - 1}{1 + \sqrt3} \end{aligned}
Why?

−π4+π3-\frac{\pi}{4} + \frac{\pi}{3} is the same as π3−π4\frac{\pi}{3} - \frac{\pi}{4}⁠, so we use the tangent of a difference:

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\begin{aligned} &\tan(A - B) \\ &= \frac{\tan A - \tan B}{1 + \tan A\tan B} \end{aligned}

with A=π3A = \frac{\pi}{3} and B=π4B = \frac{\pi}{4}⁠. The standard values are tan⁡π3=3\tan\frac{\pi}{3} = \sqrt3 and tan⁡π4=1\tan\frac{\pi}{4} = 1⁠.

tan⁡π3−tan⁡π41+tan⁡π3tan⁡π4=3−11+(3)(1)=3−11+3\begin{aligned} &\frac{\tan\frac{\pi}{3} - \tan\frac{\pi}{4}}{1 + \tan\frac{\pi}{3}\tan\frac{\pi}{4}} \\ &= \frac{\sqrt3 - 1}{1 + (\sqrt3)(1)} \\ &= \frac{\sqrt3 - 1}{1 + \sqrt3} \end{aligned}

Multiplying the numerator and the denominator by 3−1\sqrt3 - 1⁠,

3−11+3=(3−1)23−1=4−232\frac{\sqrt3 - 1}{1 + \sqrt3} = \frac{(\sqrt3 - 1)^2}{3 - 1} = \frac{4 - 2\sqrt3}{2}
Why?

This removes the root from the denominator (rationalising). Write 1+31 + \sqrt3 as 3+1\sqrt3 + 1 and multiply top and bottom by 3−1\sqrt3 - 1⁠:

3−13+1×3−13−1=(3−1)(3−1)(3+1)(3−1)\begin{aligned} &\frac{\sqrt3 - 1}{\sqrt3 + 1} \times \frac{\sqrt3 - 1}{\sqrt3 - 1} \\ &= \frac{(\sqrt3 - 1)(\sqrt3 - 1)}{(\sqrt3 + 1)(\sqrt3 - 1)} \end{aligned}

The denominator becomes a difference of squares:

(3+1)(3−1)=(3)2−(1)2=3−1=2\begin{aligned} (\sqrt3 + 1)(\sqrt3 - 1) &= (\sqrt3)^2 - (1)^2 \\ &= 3 - 1 \\ &= 2 \end{aligned}

The numerator is a perfect square:

(3−1)2=(3)2−2(3)(1)+(1)2=3−23+1=4−23\begin{aligned} (\sqrt3 - 1)^2 &= (\sqrt3)^2 - 2(\sqrt3)(1) \\ &\quad + (1)^2 \\ &= 3 - 2\sqrt3 + 1 \\ &= 4 - 2\sqrt3 \end{aligned}

Dividing both terms by 2 gives 2−32 - \sqrt3⁠.

4−232=42−232=2−3\begin{aligned} \frac{4 - 2\sqrt3}{2} &= \frac{4}{2} - \frac{2\sqrt3}{2} \\ &= 2 - \sqrt3 \end{aligned}

∴ tan⁡(tan⁡−1(−1)+π3)=2−3\therefore\ \tan\left(\tan^{-1}(-1) + \frac{\pi}{3}\right) = 2 - \sqrt3

(b)

The domain of cos⁡−1\cos^{-1} is [−1,1][-1, 1]⁠. So

Why?

cos⁡−1y\cos^{-1}y is defined only when −1≤y≤1-1 \le y \le 1⁠, because cosine never goes outside that range.

So the whole expression inside, 3x−23x - 2⁠, must lie between −1-1 and 11⁠.

−1≤3x−2≤1⇒ 1≤3x≤3⇒ 13≤x≤1\begin{aligned} &-1 \le 3x - 2 \le 1 \\ \Rightarrow\ &1 \le 3x \le 3 \\ \Rightarrow\ &\frac13 \le x \le 1 \end{aligned}
Why?

Work on all three parts at once. Add 2 to each part, then divide each part by 3.

−1+2≤3x−2+2≤1+21≤3x≤313≤3x3≤3313≤x≤1\begin{aligned} -1 + 2 \le 3x - 2 + 2 &\le 1 + 2 \\ 1 \le 3x &\le 3 \\[4pt] \frac13 \le \frac{3x}{3} &\le \frac33 \\ \frac13 \le x &\le 1 \end{aligned}

Dividing by a positive number keeps the inequality signs the same way round.

∴\therefore The domain of cos⁡−1(3x−2)\cos^{-1}(3x - 2) is [13,1]\left[\dfrac13, 1\right]⁠.

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Inverse trigonometric functions: value and domain.

Question type: (a) A trig function of an inverse + A negative number in sin⁡−1\sin^{-1}⁠, tan⁡−1\tan^{-1} or cosec⁡−1\operatorname{cosec}^{-1} (combined)

How to spot it: click to see

(a) Type 6. A trig function of an inverse: work from the inside out

Spot it: an inverse sitting inside sin, cos or tan, like cos⁡(2sin⁡−112)\cos\left(2\sin^{-1}\frac12\right)⁠.

Here: First, tan⁡−1(−1)\tan^{-1}(-1) sits inside tan: find it, then tan⁡(π3−π4)=3−11+3=2−3\tan\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\sqrt3 - 1}{1 + \sqrt3} = 2 - \sqrt3⁠.

(a) Type 2. A negative number in sin⁡−1\sin^{-1}⁠, tan⁡−1\tan^{-1} or cosec⁡−1\operatorname{cosec}^{-1}⁠: add a minus sign

Spot it: a negative number inside sin⁡−1\sin^{-1}⁠, tan⁡−1\tan^{-1} or cosec⁡−1\operatorname{cosec}^{-1}⁠.

Here: −1-1 inside tan⁡−1\tan^{-1} is negative: tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠, so tan⁡−1(−1)=−π4\tan^{-1}(-1) = -\frac{\pi}{4}⁠.

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