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NCERT Exemplar · Q47

Q.If tan⁡A=12\tan A = \dfrac{1}{2}, tan⁡B=13\tan B = \dfrac{1}{3}, then tan⁡(2A+B)\tan(2A + B) is equal to
(A) 11
(B) 22
(C) 33
(D) 44

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Compute tan⁡2A=43\tan2A=\frac{4}{3} from the double-angle formula, then apply the sum formula with tan⁡B=13\tan B=\frac13 to get tan⁡(2A+B)=3\tan(2A+B)=3 — option (C).

Step 1 — tan⁡2A\tan 2A. With tan⁡A=12\tan A=\dfrac{1}{2}:

tan⁡2A=2tan⁡A1−tan⁡2A=2⋅121−14=134=43\tan 2A=\frac{2\tan A}{1-\tan^2 A}=\frac{2\cdot\frac12}{1-\frac14}=\frac{1}{\frac34}=\frac{4}{3}

Step 2 — tan⁡(2A+B)\tan(2A+B). With tan⁡2A=43\tan 2A=\dfrac{4}{3} and tan⁡B=13\tan B=\dfrac{1}{3}: …

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