Math, asked by goyalkamlesh1402, 11 months ago

give the proper formula of tan 3 theta

Answers

Answered by durvishkhanna2006

Answer:

You can either expand or simplify the triple angle tan functions like tan 3 A, tan 3 x, tan 3 alpha etc by using triple angle identity. ... Tan 3 theta = 3 tan theta – tan 3 theta / 1 – 3 tan2 theta. Where tan is a tangent function and theta is an angle.

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Answered by pulakmath007

SOLUTION

TO DETERMINE

The formula for tan 3θ

EVALUATION

We are aware of the formula on Trigonometry that

$\displaystyle \sf \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \: \: \: - - - - (1)$

We put A = θ and B = θ then we get

$\displaystyle \sf \tan( \theta+ \theta) = \frac{\tan \theta + \tan \theta}{1 - \tan \theta \tan \theta}$

$\displaystyle \sf \implies \tan2\theta = \frac{2\tan \theta }{1 - {\tan}^{2} \theta }$

We put A = θ and B = 2θ in Equation 1

Then we get

$\displaystyle \sf \tan( \theta+2 \theta) = \frac{\tan \theta + \tan 2\theta}{1 - \tan \theta \tan 2\theta}$

$\displaystyle \sf \implies \tan 3\theta = \frac{\tan \theta + \frac{2\tan \theta }{1 - {\tan}^{2} \theta}}{1 - \tan \theta . \frac{2\tan \theta }{1 - {\tan}^{2} \theta}}$

$\displaystyle \sf \implies \tan 3\theta = \frac{ \frac{\tan \theta + 2 \tan \theta - {\tan}^{3} \theta }{1 - {\tan}^{2} \theta}}{ \frac{1 -{\tan}^{2} \theta - 2 {\tan}^{2} \theta}{1 - {\tan}^{2} \theta}}$

$\displaystyle \sf \implies \tan 3\theta = \frac{ \frac{3 \tan \theta - {\tan}^{3} \theta }{1 - {\tan}^{2} \theta}}{ \frac{1 - 3 {\tan}^{2} \theta}{1 - {\tan}^{2} \theta}}$