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
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
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 $2\cot\theta$
