Question 252 marksVery Short AnswerSection B

(a) Simplify: tan⁡−1[cos⁡2x−sin⁡2xcos⁡2x+sin⁡2x]\tan^{-1}\left[\dfrac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right], 0<x<π/40 < x < \pi/4.


OR


(b) Evaluate: tan⁡[sin⁡−11−cos⁡−1(−1/2)]\tan\left[\sin^{-1} 1 - \cos^{-1}(-1/2)\right].

Inverse Trigonometric Functions
Inverse Trigonometric Functions — Simplification and Evaluation

Official answer

Both alternatives reduce to a single standard value using inverse-trigonometric identities.


Option (a): π/4−2x\pi/4 - 2x


  • Divide numerator and denominator by cos⁡2x\cos 2x: (1−tan⁡2x)/(1+tan⁡2x)=tan⁡(π/4−2x)(1 - \tan 2x)/(1 + \tan 2x) = \tan(\pi/4 - 2x).
  • Since 0<x<π/40 < x < \pi/4, the angle π/4 − 2x lies in (−π/4,π/4)(-\pi/4, \pi/4), inside the principal branch of tan⁻¹, so tan⁡−1[tan⁡(π/4−2x)]\tan^{-1}[\tan(\pi/4 - 2x)] = π/4−2x\pi/4 - 2x.

Option (b): −1/3-1/\sqrt{3}


  • sin⁡−11=π/2\sin^{-1} 1 = \pi/2 and cos⁡−1(−1/2)=2π/3\cos^{-1}(-1/2) = 2\pi/3 (the negative argument places the angle in the second quadrant).
  • The bracket becomes π/2−2π/3=−π/6\pi/2 - 2\pi/3 = -\pi/6, so tan⁡(−π/6)\tan(-\pi/6) = −1/3-1/\sqrt{3}.
Answer keywords:inverse trigonometric functionstan(A-B) identityprincipal value branchsin inversecos inversesimplification

Marking scheme

  1. 1.1 mark: correctly reducing (a) to tan⁡(π/4−2x)\tan(\pi/4 - 2x) form, or identifying sin⁡−11=π/2\sin^{-1} 1 = \pi/2 and cos⁡−1(−1/2)=2π/3\cos^{-1}(-1/2) = 2\pi/3 in (b).
  2. 2.1 mark: correct final value — π/4−2x\pi/4 - 2x (a) with branch justification, or −1/3-1/\sqrt{3} (b).

Hint

Divide numerator and denominator of (a) by cos⁡2x\cos 2x to get tan⁡(π/4−2x)\tan(\pi/4 - 2x); for (b) recall sin⁡−11=π/2\sin^{-1} 1 = \pi/2 and cos⁡−1(−1/2)=2π/3\cos^{-1}(-1/2) = 2\pi/3.

Quick oral answer

Part (a) simplifies to π/4−2x\pi/4 - 2x via tan⁡(A−B)\tan(A-B); part (b) evaluates to −1/3-1/\sqrt{3} using sin⁡−11=π/2\sin^{-1} 1 = \pi/2 and cos⁡−1(−1/2)=2π/3\cos^{-1}(-1/2) = 2\pi/3, giving tan⁡(−π/6)\tan(-\pi/6).

Common mistakes

  1. 1.Ignoring the given domain 0<x<π/40 < x < \pi/4 and giving an angle outside the principal branch of tan⁻¹ in part (a).
  2. 2.Taking cos⁡−1(−1/2)\cos^{-1}(-1/2) as π/3\pi/3 instead of 2π/32\pi/3 — inverse cosine of a negative value lies in the second quadrant (π/2,π](\pi/2, \pi].
  3. 3.Sign slip in tan⁡(−π/6)\tan(-\pi/6), giving +1/3+1/\sqrt{3} instead of the correct −1/3-1/\sqrt{3}.

Analysis and explanation

This question tests fluency with standard inverse-trigonometric identities and principal-value ranges.


Concept: Part (a) uses the tan⁡(A−B)\tan(A - B) expansion in reverse — dividing by cos 2x converts the ratio into 11 and tan⁡2x\tan 2x. Part (b) requires the standard values sin⁡−11=π/2\sin^{-1} 1 = \pi/2 and cos⁡−1(−1/2)=2π/3\cos^{-1}(-1/2) = 2\pi/3.


Key point on (b): The inverse cosine of a negative number lies in the second quadrant (π/2,π](\pi/2, \pi], so cos⁡−1(−1/2)=2π/3\cos^{-1}(-1/2) = 2\pi/3, not π/3\pi/3.


Exam trap: In part (a) students forget to confirm the simplified angle lies within tan⁻¹'s principal branch (−π/2,π/2)(-\pi/2, \pi/2); the given domain 0<x<π/40 < x < \pi/4 exists precisely to guarantee this.


Real-world link: Such simplifications resemble phase-difference calculations in wave optics and signal processing, where phase angles are expressed as ratios of trigonometric terms.

Interesting facts

Principal value branches of inverse trigonometric functions were standardised specifically to make functions like tan⁻¹ single-valued and continuous.

The tan⁡(A−B)\tan(A-B) identity used here is the same one applied in navigation and surveying to compute bearings and angles of elevation.

Question details

Q25 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Inverse Trigonometric Functions
Topic
Inverse Trigonometric Functions — Simplification and Evaluation
Bloom's level
apply
Syllabus topic
Unit 1: Relations and Functions — Inverse Trigonometric Functions: Simplification and Standard Evaluation
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why must the domain 0<x<π/40 < x < \pi/4 be given in part (a)?

It restricts π/4−2x\pi/4 - 2x to (−π/4,π/4)(-\pi/4, \pi/4), inside tan⁻¹'s principal branch (−π/2,π/2)(-\pi/2, \pi/2), so the identity tan⁡−1[tan⁡θ]=θ\tan^{-1}[\tan \theta] = \theta applies directly without adjustment.

What are the standard values of sin⁡−11\sin^{-1} 1 and cos⁡−1(−1/2)\cos^{-1}(-1/2)?

sin⁡−11=π/2\sin^{-1} 1 = \pi/2 and cos⁡−1(−1/2)=2π/3\cos^{-1}(-1/2) = 2\pi/3. Because the argument of cos⁻¹ is negative, its value lies in the second quadrant, between π/2 and π.