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Question
If tan⁡A=12\tan A = \dfrac{1}{2} and tan⁡B=13\tan B = \dfrac{1}{3}, then tan⁡(2A+B)\tan(2A + B) is equal to:
(A)33
(B)22
(C)11
(D)32\dfrac{3}{2}
Solution Path
Double angle tan⁡2A=4/3\tan 2A = 4/3, then compound angle formula gives tan⁡(2A+B)=3\tan(2A+B) = 3.
01Question Setup
1/4
Given tan⁡A=12\tan A = \dfrac{1}{2} and tan⁡B=13\tan B = \dfrac{1}{3}. Find tan⁡(2A+B)\tan(2A + B).
tan⁡(2A+B)=  ?\tan(2A + B) = \;?
02Double Angle Formula
2/4
First compute tan⁡2A\tan 2A using tan⁡2A=2tan⁡A1−tan⁡2A=2⋅121−14=134=43\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A} = \dfrac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}} = \dfrac{1}{\frac{3}{4}} = \dfrac{4}{3}.
tan⁡2A=43\tan 2A = \dfrac{4}{3}
03Compound Angle FormulaKEY INSIGHT
3/4
Apply tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} with α=2A\alpha = 2A, β=B\beta = B. Get 43+131−49=5359=3\dfrac{\frac{4}{3} + \frac{1}{3}}{1 - \frac{4}{9}} = \dfrac{\frac{5}{3}}{\frac{5}{9}} = 3.
tan⁡(2A+B)=3\tan(2A + B) = 3
04Final Answer
4/4
tan⁡(2A+B)=3\tan(2A + B) = 3. Answer is (A).
3\boxed{3}
Concepts from this question2 concepts unlocked
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Compound Angle Expansion

EASY

Expand sin(A+B) and cos(A+B) into products of individual trig ratios of A and B

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B

The foundation of nearly all trig manipulation in JEE. Double angle, half angle, and product formulas all derive from this. Appears in 3-4 questions every paper.

Double/half angle derivationsTrig simplificationProving identities
Practice (15 Qs) →

Product-to-Sum and Sum-to-Product

STANDARD

Convert products of sin/cos into sums (and vice versa) to simplify expressions or solve equations

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A+B) + \sin(A-B)

Turns complicated products into manageable sums. Essential for integrating trig products and solving equations where factoring is needed.

Trig simplificationIntegration of trig productsSeries summation
Practice (10 Qs) →
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