Math, asked by goyalkamlesh1402, 8 months ago

give the proper formula of tan 3 theta ​

Answers

Answered by durvishkhanna2006
7

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
6

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}}

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

FINAL ANSWER

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

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