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Trigonometric Ratios for Sum of Two Angles

Prove that fo...

Question

Prove that following identities:

$t a n A + t a n (60^{\circ} + A) - t a n (60^{\circ} - A) = 3 t a n 3 A$

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Solution

$t a n A + t a n (60^{\circ} + A) - t a n (60^{\circ} - A) = 3 t a n 3 A$

$L H S = t a n A + t a n (60^{\circ} + A) - t a n (60^{\circ} - A)$

$= t a n A + \frac{t a n 60^{\circ} + t a n A}{1 - t a n 60^{\circ} t a n A} - \frac{t a n 60^{\circ} - t a n A}{1 + t a n 60^{\circ} t a n A} = t a n A + \frac{\sqrt{3} + t a n A}{1 - \sqrt{3} t a n A} - \frac{\sqrt{3} - t a n A}{1 + \sqrt{3} t a n A} = t a n A + [\frac{\sqrt{3} + 3 t a n A + t a n A + \sqrt{3} t a n^{2} A + \sqrt{3} + 3 t a n A + t a n A - \sqrt{3} t a n^{2} A}{(1 - \sqrt{3} t a n A) (1 + \sqrt{3} t a n A)}] = t a n A + \frac{8 t a n A}{1 - 3 t a n^{2} A} = \frac{t a n A - 3 t a n^{3} A + 8 t a n A}{1 - 3 t a n^{2} A} = \frac{9 t a n A - 3 t a n^{3} A}{1 - 3 t a n^{2} A} = 3 (\frac{3 t a n A - t a n^{3} A}{1 - 3 t a n^{2} A}) = 3 t a n 3 A$

So,

$t a n A + t a n (60^{\circ} + A) - t a n (60^{\circ} - A) = 3 t a n 3 A$

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\tan A + \tan (60 ° + A) - \tan (60 ° - A) = 3 \tan 3 A

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tan A + tan (60^{\circ} + A) + tan (120^{\circ} + A) = 3 tan 3 A

tan 60^{\circ} tan 80^{\circ} tan 20^{\circ} = tan 80^{\circ} - tan 20^{\circ} - tan 60^{\circ}

tan 60^{\circ} = \frac{2 tan 30^{\circ}}{1 - {tan}^{2} 30^{\circ}}

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Trigonometric Ratios for Sum of Two Angles

Standard XII Mathematics

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