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The expression

$\tan\theta+2\tan(2\theta)+2^2\tan(2^2\theta)+\dots+2^{14}\tan(2^{14}\theta)+2^{15}\cot(2^{15}\theta)$ equals to :

The answer in the answer book is given to be $\cot\theta$. I am unable to think in which way to proceed to find the required solution.

thanks in advance!

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    $\begingroup$ Look at $\tan x + 2 \cot (2x)$, and see whether you get something telescoping. $\endgroup$ – Daniel Fischer May 2 '15 at 9:41
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HINT :

$$a\tan\beta+2a\cot(2\beta)=a\left(\tan\beta+2\cdot\frac{1-\tan^2\beta}{2\tan\beta}\right)=a\cot\beta.$$ So, one has $$2^{14}\tan(2^{14}\theta)+2^{15}\cot(2^{15}\theta)=2^{14}\cot(2^{14}\theta)$$ $$2^{13}\tan(2^{13}\theta)+2^{14}\cot(2^{14}\theta)=2^{13}\cot(2^{13}\theta)$$

and so on.

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$$\cot x-\tan x=\dfrac{\cos^2x-\sin^2x}{\sin x\cos x}=2\cot2x$$

Set $x=\theta,2\theta,2^2\theta,\cdots.2^n\theta$ where $n$ is a non-negative integer

$$\cot\theta-\tan\theta=2\cot2\theta\ \ \ \ (1)$$

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

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

$$\cdots$$

$$2^n(\cot2^n\theta-\tan2^n\theta)=2^n(2\cot2^{n+1}\theta)\ \ \ \ (n)$$

Adding we get $$\displaystyle\cot\theta-\sum_{r=1}^n2^r\tan2^r\theta=2^{n+1}\cot2^{n+1}\theta$$

$$\displaystyle\iff\sum_{r=1}^n2^r\tan2^r\theta=\cot\theta-2^{n+1}\cot2^{n+1}\theta$$

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