<s>[INST] 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
 [/INST] 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
Now add the relations.
Reference : Double-Angle Formulas
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