Question 313 marksShort AnswerSection C

  1. (B) Prove that : tan⁡A1+sec⁡A−tan⁡A1−sec⁡A=2csc⁡A\dfrac{\tan A}{1 + \sec A} - \dfrac{\tan A}{1 - \sec A} = 2\csc A.

Introduction to Trigonometry
Proving a trigonometric identity

Official answer

Proved: LHS=2csc⁡A=RHSLHS = 2\csc A = RHS. Taking tan A common and combining the two fractions over the common denominator (1+sec⁡A)(1−sec⁡A)=1−sec⁡2A=−tan⁡2A(1 + \sec A)(1 - \sec A) = 1 - \sec^2 A = -\tan^2 A (using 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A), the LHS reduces to 2sec⁡A/tan⁡A2\sec A/\tan A. Converting to sin A and cos A gives 2/sin⁡A=2csc⁡A2/\sin A = 2\csc A, which equals the RHS. Hence the identity is established.

Answer keywords:trigonometric identitytan Asec Acosec A1 + tan²A = sec²Acommon denominatordifference of squaresprove

Marking scheme

  1. 1.1 mark: Taking tan A common and correctly combining the two fractions over the denominator (1+sec⁡A)(1−sec⁡A)(1 + \sec A)(1 - \sec A).
  2. 2.1 mark: Simplifying the denominator to 1−sec⁡2A1 - \sec^2 A and correctly replacing it with −tan⁡2A-\tan^2 A (the sign must be handled correctly).
  3. 3.1 mark: Reducing 2sec⁡A/tan⁡A2\sec A / \tan A to 2/sin⁡A=2csc⁡A2/\sin A = 2\csc A and concluding LHS=RHSLHS = RHS. Accept any valid alternative route (e.g., converting everything to sin A and cos A from the start).

Hint

Take tan A common first, combine over the common denominator (1+sec⁡A)(1−sec⁡A)(1+\sec A)(1-\sec A), and remember that 1−sec⁡2A=−tan⁡2A1 - \sec^2 A = -\tan^2 A.

Quick oral answer

I take tan A common, combine the two fractions so the denominator becomes 1−sec⁡2A1 - \sec^2 A which equals −tan⁡2A-\tan^2 A, giving 2sec⁡A2\sec A over tan⁡A\tan A, and that reduces to 2 by sin A, i.e., 2csc⁡A2\csc A.

Common mistakes

  1. 1.Writing 1−sec⁡2A=tan⁡2A1 - \sec^2 A = \tan^2 A (missing the negative sign), which produces −2csc⁡A-2\csc A instead of 2csc⁡A2\csc A.
  2. 2.Trying to prove the result by cross-multiplying and manipulating both sides at once instead of transforming only the LHS — CBSE penalises 'working from the answer'.
  3. 3.Errors while combining the fractions, especially sign slips in the numerator (1−sec⁡A)−(1+sec⁡A)=−2sec⁡A(1 - \sec A) - (1 + \sec A) = -2\sec A.

Analysis and explanation

A classic 'simplify one side' trigonometric identity built on the Pythagorean relation 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A.


Concept & strategy

  • Take tan A common from both terms on the LHS to avoid expanding messy products.
  • Combine over the common denominator (1+sec⁡A)(1−sec⁡A)=1−sec⁡2A(1 + \sec A)(1 - \sec A) = 1 - \sec^2 A, then rewrite this as −tan⁡2A-\tan^2 A.
  • Convert the final ratio sec⁡A/tan⁡A\sec A/\tan A into sin A, cos A to land on cosec A.

Common mistakes

  • Writing 1−sec⁡2A=tan⁡2A1 - \sec^2 A = \tan^2 A (dropping the negative sign) — this is where most students lose marks and end up with −2csc⁡A-2\csc A instead of +2csc⁡A+2\csc A.
  • Manipulating both sides of the identity simultaneously instead of transforming only the LHS — CBSE examiners deduct marks for this 'working backward' approach.

Real-world relevance

  • Interconversions between reciprocal trigonometric ratios like this are routine in resolving forces in physics, analysing AC waveforms in electrical engineering, and surveying calculations.

Interesting facts

The identity 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A is a direct algebraic consequence of dividing the fundamental identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1 by cos²A — every 'sec/tan' identity in the CBSE syllabus traces back to this one Pythagorean relation.

The reciprocal ratios cosec, sec and cot were historically tabulated by astronomers; the Indian mathematician Aryabhata (c. 499 CE) computed sine tables (jya) that European trigonometry later built upon.

Expressions of the form 1/(1+sec⁡A)−1/(1−sec⁡A)1/(1+\sec A) - 1/(1-\sec A) appear in electrical engineering when combining reactances, where the difference-of-squares simplification saves substantial computation.

Question details

Q31 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Introduction to Trigonometry
Topic
Proving a trigonometric identity
Bloom's level
apply
Syllabus topic
Unit 3: Trigonometry — Introduction to Trigonometry: Trigonometric Identities (CBSE Class 10 Mathematics)
Author
CBSE
Reviewed by
Padhantu Academic Team

Spotted a mistake or something unclear?

Tell us and we will fix it.

Frequently asked questions

Why does 1−sec⁡2A1 - \sec^2 A become −tan⁡2A-\tan^2 A?

From the identity 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A, rearranging gives sec⁡2A−1=tan⁡2A\sec^2 A - 1 = \tan^2 A. Therefore 1−sec⁡2A=−(sec⁡2A−1)=−tan⁡2A1 - \sec^2 A = -(\sec^2 A - 1) = -\tan^2 A. The negative sign is essential; dropping it flips the final sign of the answer.

Can I solve this by converting everything to sin A and cos A at the start?

Yes. Replace tan⁡A\tan A with sin⁡A/cos⁡A\sin A/\cos A and sec⁡A\sec A with 1/cos⁡A1/\cos A, simplify each fraction, and combine. You will reach 2/sin⁡A=2csc⁡A2/\sin A = 2\csc A. It is a valid route and earns full marks, though taking tan A common is usually quicker.

Is it acceptable to prove RHS = LHS instead of LHS = RHS?

You may start from either side and simplify toward the other, but you must transform only one side at a time. What CBSE penalises is manipulating both sides simultaneously (cross-multiplying the whole equation), which assumes the result you are asked to prove.