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
 A: 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
A: 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 $2\cot\theta$
