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CBSE Board Examination 2026 · Section B · 2 marks

Q25: Simplify tan⁡−1cos⁡2x−sin⁡2xcos⁡2x+sin⁡2x\tan^{-1}\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}⁠, or evaluate tan⁡(sin⁡−11−cos⁡−1(−12))\tan\left(\sin^{-1}1 - \cos^{-1}\left(-\frac12\right)\right)

Board 2026Ch 2 · Inverse Trigonometric Functions
The question

(a) Simplify :

tan⁡−1(cos⁡2x−sin⁡2xcos⁡2x+sin⁡2x),\tan^{-1}\left(\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right),

0<x<π40 < x < \dfrac{\pi}{4}⁠.

OR

(b) Evaluate : tan⁡(sin⁡−11−cos⁡−1(−12))\tan\left(\sin^{-1}1 - \cos^{-1}\left(-\dfrac12\right)\right)

Exam answer

(a)

As 0<2x<π20 < 2x < \frac{\pi}{2}⁠, cos⁡2x>0\cos 2x > 0⁠. Dividing the numerator and the denominator by cos⁡2x\cos 2x⁠,

tan⁡−1(cos⁡2x−sin⁡2xcos⁡2x+sin⁡2x)=tan⁡−1(1−tan⁡2x1+tan⁡2x)\begin{aligned} &\tan^{-1}\left(\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right) \\ &= \tan^{-1}\left(\frac{1 - \tan 2x}{1 + \tan 2x}\right) \end{aligned}
Why?

Dividing the top and the bottom by cos⁡2x\cos 2x turns sin⁡2xcos⁡2x\frac{\sin 2x}{\cos 2x} into tan⁡2x\tan 2x and cos⁡2xcos⁡2x\frac{\cos 2x}{\cos 2x} into 1. The value of the fraction does not change.

cos⁡2x−sin⁡2xcos⁡2x+sin⁡2x=cos⁡2xcos⁡2x−sin⁡2xcos⁡2xcos⁡2xcos⁡2x+sin⁡2xcos⁡2x=1−tan⁡2x1+tan⁡2x\begin{aligned} &\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x} \\ &= \frac{\frac{\cos 2x}{\cos 2x} - \frac{\sin 2x}{\cos 2x}}{\frac{\cos 2x}{\cos 2x} + \frac{\sin 2x}{\cos 2x}} \\ &= \frac{1 - \tan 2x}{1 + \tan 2x} \end{aligned}

This is allowed because cos⁡2x≠0\cos 2x \ne 0⁠: for 0<x<π40 < x < \frac{\pi}{4}⁠, the angle 2x2x is between 0 and π2\frac{\pi}{2}⁠.

Using 1−tan⁡A1+tan⁡A=tan⁡(π4−A)\frac{1 - \tan A}{1 + \tan A} = \tan\left(\frac{\pi}{4} - A\right)⁠,

=tan⁡−1[tan⁡(π4−2x)]= \tan^{-1}\left[\tan\left(\frac{\pi}{4} - 2x\right)\right]
Why?

This comes from the formula for tan⁡(A−B)\tan(A - B) with tan⁡π4=1\tan\frac{\pi}{4} = 1⁠:

tan⁡(π4−A)=tan⁡π4−tan⁡A1+tan⁡π4tan⁡A=1−tan⁡A1+tan⁡A\begin{aligned} \tan\left(\frac{\pi}{4} - A\right) &= \frac{\tan\frac{\pi}{4} - \tan A}{1 + \tan\frac{\pi}{4}\tan A} \\ &= \frac{1 - \tan A}{1 + \tan A} \end{aligned}

Here A=2xA = 2x⁠.

Since 0<x<π40 < x < \frac{\pi}{4}⁠, −π4<π4−2x<π4-\frac{\pi}{4} < \frac{\pi}{4} - 2x < \frac{\pi}{4}⁠, which lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠.

Why?

tan⁡−1(tan⁡θ)=θ\tan^{-1}(\tan\theta) = \theta only when θ\theta is in the principal value range of tan⁡−1\tan^{-1}⁠, which is (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠.

From 0<x<π40 < x < \frac{\pi}{4}⁠: multiply by −2-2 (the signs flip) and add π4\frac{\pi}{4} to every part.

0<x<π40>−2x>−π2π4+0>π4−2x>π4−π2π4>π4−2x>−π4\begin{aligned} 0 &< x < \frac{\pi}{4} \\ 0 &> -2x > -\frac{\pi}{2} \\ \frac{\pi}{4} + 0 &> \frac{\pi}{4} - 2x > \frac{\pi}{4} - \frac{\pi}{2} \\ \frac{\pi}{4} &> \frac{\pi}{4} - 2x > -\frac{\pi}{4} \end{aligned}

(π4−π2=π4−2π4=−π4\frac{\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4} - \frac{2\pi}{4} = -\frac{\pi}{4}⁠.)

Watch out: always check this range before cancelling tan⁡−1\tan^{-1} with tan⁡\tan⁠.

∴\therefore The given expression =π4−2x= \frac{\pi}{4} - 2x⁠.

(b)

sin⁡−11=π2\sin^{-1}1 = \frac{\pi}{2} and cos⁡−1(−12)=π−π3=2π3\cos^{-1}\left(-\frac12\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}⁠.

Why?

Principal values: sin⁡−1\sin^{-1} gives an angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, and cos⁡−1\cos^{-1} gives an angle in [0,π][0, \pi]⁠.

sin⁡π2=1\sin\frac{\pi}{2} = 1⁠, so sin⁡−11=π2\sin^{-1}1 = \frac{\pi}{2}⁠.

For the negative value, use

cos⁡−1(−a)=π−cos⁡−1a\cos^{-1}(-a) = \pi - \cos^{-1}a

with cos⁡−112=π3\cos^{-1}\frac12 = \frac{\pi}{3}⁠. Watch out: −π3-\frac{\pi}{3} is not in [0,π][0, \pi]⁠, so it is not the answer.

cos⁡−1(−12)=π−cos⁡−112=π−π3=3π3−π3=2π3\begin{aligned} \cos^{-1}\left(-\tfrac12\right) &= \pi - \cos^{-1}\tfrac12 \\ &= \pi - \frac{\pi}{3} \\ &= \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3} \end{aligned}
tan⁡(sin⁡−11−cos⁡−1(−12))=tan⁡(π2−2π3)=tan⁡(−π6)=−tan⁡π6\begin{aligned} &\tan\left(\sin^{-1}1 - \cos^{-1}\left(-\frac12\right)\right) \\ &= \tan\left(\frac{\pi}{2} - \frac{2\pi}{3}\right) \\ &= \tan\left(-\frac{\pi}{6}\right) = -\tan\frac{\pi}{6} \end{aligned}
Why?

Over the common denominator 6,

π2=3π6,2π3=4π6π2−2π3=3π−4π6=−π6\begin{aligned} \frac{\pi}{2} &= \frac{3\pi}{6},\quad \frac{2\pi}{3} = \frac{4\pi}{6} \\[4pt] \frac{\pi}{2} - \frac{2\pi}{3} &= \frac{3\pi - 4\pi}{6} = -\frac{\pi}{6} \end{aligned}

tan⁡\tan is an odd function, tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta⁠, and tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt3}⁠.

∴\therefore The given expression =−13= -\dfrac{1}{\sqrt3}⁠.

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Inverse trig functions: simplify, evaluate.

Question type: (b) A trig function of an inverse + A negative number in cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} or cot⁡−1\cot^{-1} (combined)

How to spot it: click to see

(b) 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, the two inverses sit inside tan: find each one, so the bracket is π2−2π3=−π6\frac{\pi}{2} - \frac{2\pi}{3} = -\frac{\pi}{6}⁠, and only then take tan.

(b) Type 3. A negative number in cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} or cot⁡−1\cot^{-1}⁠: take it away from π\pi

Spot it: a negative number inside cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} or cot⁡−1\cot^{-1}⁠.

Here: cos⁡−1(−12)\cos^{-1}\left(-\frac12\right) has a negative number: it is π−π3=2π3\pi - \frac{\pi}{3} = \frac{2\pi}{3}⁠, not −π3-\frac{\pi}{3}⁠.

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