OR

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

Q: OR Prove that : $\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2\csc A$

$LHS = \frac{\tan A}{1+\sec A} - \frac{\tan A}{1-\sec A}$. Taking LCM: = $\frac{\tan A[(1-\sec A)-(1+\sec A)]}{1-\sec^2 A} = \frac{-2\tan A \sec A}{-\tan^2 A} = \frac{2\sec A}{\tan A} = 2 \times \frac{1}{\cos A} \times \frac{\cos A}{\sin A} = \frac{2}{\sin A} = 2\csc A = RHS$. Hence proved.

Q: Which trigonometric identity is used to solve this?

The Pythagorean identity $1 + \tan^2 A = \sec^2 A$, rearranged as $\sec^2 A - 1 = \tan^2 A$, is the key identity used to simplify the denominator $(1+\sec A)(1-\sec A) = 1 - \sec^2 A = -\tan^2 A$.

Introduction to Trigonometry

Trigonometric Identities

Official Answer

$LHS = \frac{\tan A}{1+\sec A} - \frac{\tan A}{1-\sec A}$ . Taking LCM: = $\frac{\tan A[(1-\sec A)-(1+\sec A)]}{1-\sec^2 A} = \frac{-2\tan A \sec A}{-\tan^2 A} = \frac{2\sec A}{\tan A} = 2 \times \frac{1}{\cos A} \times \frac{\cos A}{\sin A} = \frac{2}{\sin A} = 2\csc A = RHS$ . Hence proved.

trigonometric identitytan Asec Acosec A1+tan²A=sec²ALHS=RHSalgebraic simplification

Marking Scheme

11 mark: correctly taking LCM of the two fractions and simplifying the numerator $\tan A[(1-\sec A)-(1+\sec A)] = -2\tan A \sec A$ .
21 mark: simplifying the denominator $(1+\sec A)(1-\sec A) = 1 - \sec^2 A = -\tan^2 A$ .
31 mark: final simplification to $\frac{2\sec A}{\tan A} = 2\csc A$ , matching RHS, with clear concluding statement 'Hence Proved'.

Hint

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

Quick Oral Answer

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

Analysis & Explanation

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

Concept

Uses the Pythagorean identity $1 + \tan^2 A = \sec^2 A$ , rearranged as $\sec^2 A - 1 = \tan^2 A$ .
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: $(1+\sec A)(1-\sec A) = 1 - \sec^2 A$ .
Convert $\frac{\sec A}{\tan A}$ to cos and sin form to reach cosec A.

Common mistakes

Losing the negative sign when writing $1 - \sec^2 A = -\tan^2 A$ , 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.

Common Mistakes

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

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

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Frequently Asked Questions

Which trigonometric identity is used to solve this?

The Pythagorean identity $1 + \tan^2 A = \sec^2 A$ , rearranged as $\sec^2 A - 1 = \tan^2 A$ , is the key identity used to simplify the denominator $(1+\sec A)(1-\sec A) = 1 - \sec^2 A = -\tan^2 A$ .

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.