Prove that

\(\tan n\theta =\large{ \frac{ \binom{n}{1}t - \binom{n}{3}t^{3} + \binom{n}{5}t^{5} - ...........}{ 1 - \binom{n}{2}t^{2} + \binom{n}{4}t^{4} - .........................}}\)

where \(t = \tan \theta\)

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## Comments

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TopNewestBy De Moivre's theorem

\((cos\theta + i sin\theta)^{n} = cos(n\theta) + i sin(n\theta)\)

Writting the binomial expression of

\((cos\theta + i sin\theta)^{n}\)

Now

Equating real part to

\(cos(n\theta)\)And imaginary part to

\(sin(n\theta)\)We get(let cos = c, sin = s)

\(\displaystyle c(n\theta) = c^{n} - C_2^{n} c^{n -2} s^{2} +\ldots\)

\(\displaystyle s(n\theta) = C_1^{n} c^{n-1} s - C_3^{n}c^{n-3} s^{3} + \ldots\)

Dividing equation 2 by 1 we get

\(\displaystyle t(n\theta) = \frac{C_1^{n} c^{n-1}s - C_3^{n} c^{n-3}s^{3} + \ldots}{c^{n} - C_2^{n}c^{n-2}s^{2} + \ldots}\)

Now divide by \(\cos^{n} \theta\) in numerator and denominator to get the required expression.

Hence Proved!

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Thanks Good solution

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@Calvin Lin @DEEPANSHU GUPTA @Kostub Deshmukh @Agnishom Chattopadhyay @Pratik Shastri @Pranshu Gaba @Pranav Arora @Pranjal Jain @Victor Loh @Vinay Sipani @Aditya Raut @Christopher Boo @Krishna Ar @Krishna Sharma

Please help

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@brian charlesworth @Satvik Golechha @Mvs Saketh @Santanu Banerjee @Aman Sharma

you too please help

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@Sanjeet Raria You too please

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@Mursalin Habib @John Muradeli @Cody Johnson You too please help

To Brilliant.org ,

Here after a comment is posted and when we click the edit button and want mention someone the menu is not coming . we have to make a new comment for this

For example firstly I forget about the people in the second, third and four comment and when i clicked the edit and type their name the menu was not displayed

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This does need to be fixed, though. I thought I was the only one, but apparently not.

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I would solve this one by using De Moiver's identity (cosθ+jsinθ)^n=cosnθ+jsinnθ. Then tannθ=sinnθ/cosnθ. By expanding into powers we have (cosθ+jsinθ)^n=Sum(j^k *sin(θ)^k * cos(θ)^(n-k) * (n per k) )=Sum(j^k *tan(θ)^k * cos(θ)^n * (n per k) ). By grouping into real and imaginary parts we can get the result given. (I should learn to write in these posts in a more beautiful way :))) )

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thanks for proving, refer this

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Thanks!!

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tan (A+B+C+......)= \(\frac{S_{1}-S_{3}+S_{5}-.....}{1-S_{2}+S_{4}-......}\)Where \(S_{i}\) is sum of products of i terms taken at a time.

For A=B=C...=\(\theta\), You will get desired result!

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