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Trigonometric Ratios for Sum of Two Angles

If tan A+B=p,...

Question

If tan(A+B)=p, tan(A-B)=q, then show that $t a n 2 A = \frac{p + q}{1 - p q}$ .

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Solution

RHS,
$\frac{p + q}{1 - p q} = \frac{t a n (A + B) + t a n (A - B)}{1 - t a n (A + B) . t a n (A - B)}$
$= \frac{\frac{t a n A + t a n B}{1 - t a n A . t a n B} + \frac{t a n A - t a n B}{1 + t a n A . t a n B}}{1 - \frac{t a n A + t a n B}{1 - t a n A . t a n B} . \frac{t a n A - t a n B}{1 + t a n A . t a n B}}$
$= \frac{\frac{(t a n A + t a n B) (1 + t a n A . t a n B) + (t a n A - t a n B)}{(1 - t a n A . t a n B) (1 + t a n A . t a n B)}}{\frac{(1 - t a n A . t a n B) (1 + t a n A . t a n B) + (t a n A + t a n B) . (t a n A - t a n B)}{(1 - t a n A . t a n B) (1 + t a n A . t a n B)}}$
$= \frac{t a n A + t a n B + t a n^{2} A . t a n B + t a n A . t a n^{2} B + t a n A - t a n B - t a n^{2} A . t a n B + t a n A . t a n^{2} B}{1 - t a n^{2} A . t a n^{2} B - t a n^{2} A + t a n^{2} B}$
$= \frac{2 t a n A + 2 t a n A . t a n^{2} B}{(1 - t a n^{2} A) (1 + t a n^{2} B)} = \frac{2 t a n A (1 + t a n^{2} B)}{(1 - t a n^{2} A) (1 + t a n^{2} B)} = \frac{2 t a n A}{1 - t a n^{2} A} = t a n 2 A = L H S$
Hence proved.

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Trigonometric Ratios for Sum of Two Angles

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