# Simplify the expression : $\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\cdots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$

How to simplify the expression:

$\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\ldots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$

I am not getting any clue how to proceed in such problem please suggest it will be of great help .. I got this problem from www.mathstudy.in

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HINT:

Use $$\cot A-\tan A=\frac{\cos^2A-\sin^2A}{\cos A\sin A}=2\cot2A$$ repeatedly

So, we have $$\cot\theta-\tan\theta=2\cot2 \theta$$

and $$2(\cot2\theta-\tan2\theta)=2(2\cot2^2\theta)$$

$$2^2(\cot2^2\theta-\tan2^2\theta)=2^2(2\cot2^3\theta)$$ and so on

Reference : Double-Angle Formulas

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We have to find the value of $\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\ldots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$ .

$2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15}\theta)$=$2^{14} \tan(2^{14}\theta) +2^{15} \dfrac{1}{\tan(2^{15}\theta)}$ = $2^{14} \tan(2^{14}\theta) +2^{15} \dfrac{1-\tan^2(2^{14}\theta)}{2\tan(2^{14}\theta)}$ = $\dfrac{2^{15}}{2\tan(2^{14}\theta)}$ = $2^{14}cot(2^{14}\theta)$

By this you can prove that $2^{13} \tan(2^{13}\theta)+2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15}\theta)$=$2^{13}cot(2^{13}\theta)$

hence you can prove that By this you can prove that $2^{12} \tan(2^{12}\theta)+2^{13} \tan(2^{13}\theta)+2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15}\theta)$=$2^{12}cot(2^{12}\theta)$

So the result of $\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\ldots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$ is $2\cot\theta$

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