Question 313 marksShort AnswerSection C

OR

Prove that : tan⁡A1+sec⁡A−tan⁡A1−sec⁡A=2csc⁡A\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2\csc A

Introduction to Trigonometry
Trigonometric Identities

Official answer

LHS=tan⁡A1+sec⁡A−tan⁡A1−sec⁡ALHS = \frac{\tan A}{1+\sec A} - \frac{\tan A}{1-\sec A}. Taking LCM: = tan⁡A[(1−sec⁡A)−(1+sec⁡A)]1−sec⁡2A=−2tan⁡Asec⁡A−tan⁡2A=2sec⁡Atan⁡A=2×1cos⁡A×cos⁡Asin⁡A=2sin⁡A=2csc⁡A=RHS\frac{\tan A[(1-\sec A)-(1+\sec A)]}{1-\sec^2 A} = \frac{-2\tan A \sec A}{-\tan^2 A} = \frac{2\sec A}{\tan A} = 2 \times \frac{1}{\cos A} \times \frac{\cos A}{\sin A} = \frac{2}{\sin A} = 2\csc A = RHS. Hence proved.

Answer keywords:trigonometric identitytan Asec Acosec A1+tan²A=sec²ALHS=RHSalgebraic simplification

Marking scheme

  1. 1.1 mark: correctly taking LCM of the two fractions and simplifying the numerator tan⁡A[(1−sec⁡A)−(1+sec⁡A)]=−2tan⁡Asec⁡A\tan A[(1-\sec A)-(1+\sec A)] = -2\tan A \sec A.
  2. 2.1 mark: simplifying the 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.
  3. 3.1 mark: final simplification to 2sec⁡Atan⁡A=2csc⁡A\frac{2\sec A}{\tan A} = 2\csc A, matching RHS, with clear concluding statement 'Hence Proved'.

Hint

Take LCM of the two fractions on the LHS and use 1−sec⁡2A=−tan⁡2A1 - \sec^2 A = -\tan^2 A to simplify.

Quick oral answer

Take LCM of the two LHS fractions, use 1−sec⁡2A=−tan⁡2A1 - \sec^2 A = -\tan^2 A to simplify the denominator, cancel tan A, and reduce sec⁡Atan⁡A\frac{\sec A}{\tan A} to cosec A to match the RHS.

Common mistakes

  1. 1.Sign error when simplifying 1−sec⁡2A1 - \sec^2 A as tan²A instead of −tan²A, flipping the final sign of the answer.
  2. 2.Cross-multiplying the equation as if solving for A instead of only manipulating the LHS to reach the RHS.
  3. 3.Writing tan⁡Asec⁡A=sin⁡A\frac{\tan A}{\sec A} = \sin A instead of correctly simplifying sec⁡Atan⁡A=csc⁡A\frac{\sec A}{\tan A} = \csc A at the final step.

Analysis and explanation

Proving a trigonometric identity by combining fractions over a common denominator and applying sec⁡2A−1=tan⁡2A\sec^2 A - 1 = \tan^2 A.


Concept

  • Uses the Pythagorean identity 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A, rearranged as sec⁡2A−1=tan⁡2A\sec^2 A - 1 = \tan^2 A.
  • LHS must be simplified independently until it matches RHS — never cross-multiply across the '=' sign.

Key steps

  • Combine the two fractions using difference of squares: (1+sec⁡A)(1−sec⁡A)=1−sec⁡2A(1+\sec A)(1-\sec A) = 1 - \sec^2 A.
  • Convert sec⁡Atan⁡A\frac{\sec A}{\tan A} to cos and sin form to reach cosec A.

Common mistakes

  • Losing the negative sign when writing 1−sec⁡2A=−tan⁡2A1 - \sec^2 A = -\tan^2 A, which leads to −2 cosec A instead of +2 cosec A.
  • Attempting to cross-multiply the identity instead of simplifying LHS alone.

Real-world

  • Such trigonometric simplifications underpin wave, optics and engineering calculations, though here the goal is purely algebraic fluency for the CBSE board exam.

Interesting facts

The word 'trigonometry' comes from Greek 'trigonon' (triangle) and 'metron' (measure), and such identities were used by ancient astronomers like Hipparchus to build the first trigonometric tables around 150 BCE.

CBSE class 10 boards have asked a 'prove the trigonometric identity' question in almost every year's paper since 2015, making this one of the most predictable 3-mark question types.

The identity 1+tan⁡2A=sec⁡2A1+\tan^2 A = \sec^2 A used here is derived directly from sin⁡2A+cos⁡2A=1\sin^2 A+\cos^2 A=1 by dividing throughout by cos²A.

Question details

Q31 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Introduction to Trigonometry
Topic
Trigonometric Identities
Bloom's level
apply
Syllabus topic
Unit 5: Trigonometry — Introduction to Trigonometry: Trigonometric Identities
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Which trigonometric identity is used to solve this?

The Pythagorean identity 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A, rearranged as sec⁡2A−1=tan⁡2A\sec^2 A - 1 = \tan^2 A, is the key identity used to simplify the 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.

Can this identity be proved by starting from the RHS instead?

Yes, but CBSE convention and simplicity favour starting from the more complex side (LHS with two fractions) and simplifying down to the RHS; starting from RHS and expanding backward is harder to justify step-by-step.