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NCERT Exemplar · Q11

Q.If tan⁡(A+B)=p\tan(A + B) = p, tan⁡(A−B)=q\tan(A - B) = q, then show that tan⁡2A=p+q1−pq\tan 2A = \dfrac{p + q}{1 - pq}.

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The key idea is to express 2A2A as the sum (A+B)+(A−B)(A+B)+(A-B), then apply the tangent addition formula. This directly gives tan⁡2A=p+q1−pq\tan 2A = \frac{p+q}{1-pq}.

We are given tan⁡(A+B)=p\tan(A+B) = p and tan⁡(A−B)=q\tan(A-B) = q. The goal is to find tan⁡2A\tan 2A in terms of pp and qq.

The natural instinct is to notice that 2A2A can be written as the sum of the two given angles:

2A=(A+B)+(A−B)2A = (A+B) + (A-B)

This is the crucial observation. Once we see this, the problem reduces to applying a standard trigonometric identity — the tangent addition formula.

tan⁡(X+Y)=tan⁡X+tan⁡Y1−tan⁡Xtan⁡Y\tan(X+Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y}

Let’s apply it step by step.

  1. Set up the identity. Let X=A+BX = A+B and Y=A−BY = A-B. Then:

tan⁡2A=tan⁡((A+B)+(A−B))=tan⁡(A+B)+tan⁡(A−B)1−tan⁡(A+B)tan⁡(A−B)\tan 2A = \tan\big((A+B)+(A-B)\big) = \frac{\tan(A+B) + \tan(A-B)}{1 - \tan(A+B)\tan(A-B)}

  1. Substitute the given values. We know tan⁡(A+B)=p\tan(A+B) = p and tan⁡(A−B)=q\tan(A-B) = q. So:

tan⁡2A=p+q1−p⋅q\tan 2A = \frac{p + q}{1 - p \cdot q}

  1. Simplify. The expression is already in its simplest form: tan⁡2A=p+q1−pq\tan 2A = \frac{p+q}{1-pq} …

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