Q.If $\tan\alpha = \dfrac{1}{3}$ and $\tan\beta = \dfrac{1}{7}$ then prove that $(2\alpha + \beta) = \dfrac{\pi}{4}$.

Compute $\tan2\alpha=\dfrac34$ from the double-angle formula, then $\tan(2\alpha+\beta)=1$ from the addition formula, giving $2\alpha+\beta=\dfrac{\pi}{4}$.

A compound-angle proof from the Trigonometry unit of the Tamil Nadu HSC Class-11 Business Mathematics syllabus.

Step 1 — Find $\tan2\alpha$. With $\tan\alpha=\dfrac13$,

$\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{2\cdot\tfrac13}{1-\tfrac19}=\frac{\tfrac23}{\tfrac89}=\frac{2}{3}\cdot\frac{9}{8}=\frac{3}{4}.$

Step 2 — Apply the addition formula with $\tan\beta=\dfrac17$.

$\tan(2\alpha+\beta)=\frac{\tan2\alpha+\tan\beta}{1-\tan2\alpha\tan\beta}=\frac{\tfrac34+\tfrac17}{1-\tfrac34\cdot\tfrac17}.$

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