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

Q: 28. (B) Prove that : $\dfrac{\tan A}{1 + \sec A} - \dfrac{\tan A}{1 - \sec A} = 2\csc A$.

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

Q: Why does $1 - \sec^2 A$ become $-\tan^2 A$?

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

Introduction to Trigonometry

Proving a trigonometric identity

Official Answer

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

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

Marking Scheme

11 mark: Taking tan A common and correctly combining the two fractions over the denominator $(1 + \sec A)(1 - \sec A)$ .
21 mark: Simplifying the denominator to $1 - \sec^2 A$ and correctly replacing it with $-\tan^2 A$ (the sign must be handled correctly).
31 mark: Reducing $2\sec A / \tan A$ to $2/\sin A = 2\csc A$ and concluding $LHS = RHS$ . 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 $(1+\sec A)(1-\sec A)$ , and remember that $1 - \sec^2 A = -\tan^2 A$ .

Quick Oral Answer

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

Analysis & Explanation

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

Concept & strategy

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

Common mistakes

Writing $1 - \sec^2 A = \tan^2 A$ (dropping the negative sign) — this is where most students lose marks and end up with $-2\csc A$ instead of $+2\csc A$ .
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.

Common Mistakes

1Writing $1 - \sec^2 A = \tan^2 A$ (missing the negative sign), which produces $-2\csc A$ instead of $2\csc A$ .
2Trying 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'.
3Errors while combining the fractions, especially sign slips in the numerator $(1 - \sec A) - (1 + \sec A) = -2\sec A$ .

Interesting Facts

The identity $1 + \tan^2 A = \sec^2 A$ is a direct algebraic consequence of dividing the fundamental identity $\sin^2 A + \cos^2 A = 1$ 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 $1/(1+\sec A) - 1/(1-\sec A)$ appear in electrical engineering when combining reactances, where the difference-of-squares simplification saves substantial computation.

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Frequently Asked Questions

Why does $1 - \sec^2 A$ become $-\tan^2 A$ ?

From the identity $1 + \tan^2 A = \sec^2 A$ , rearranging gives $\sec^2 A - 1 = \tan^2 A$ . Therefore $1 - \sec^2 A = -(\sec^2 A - 1) = -\tan^2 A$ . 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 $\tan A$ with $\sin A/\cos A$ and $\sec A$ with $1/\cos A$ , simplify each fraction, and combine. You will reach $2/\sin A = 2\csc A$ . 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.