Question

$\tan \alpha +2\tan 2\alpha +2^2\tan 2^2\alpha +2^3\tan 2^3\alpha +\cdots +2^n\tan 2^n+2^{n+1}\cot 2^{n+1}\alpha$ equals $\forall n\in N$.

• A. $\tan \alpha$
• B. $\tan \alpha -\cot \alpha$
• C. $\cot \alpha$
• D. None of these

Answer 1

Answer: (C)

$\cot \alpha$

$cot\theta -\tan \theta =\frac{1+{\tan }^2\theta }{\tan \theta }=2\left(\frac{1-{\tan }^2\theta }{2\tan \theta }\right)=2cot2\theta$
Now
$\tan \alpha +2\tan 2\alpha +2^2\tan 2^2\alpha +...2^n\tan 2^n\alpha +2^{n+1}cot{2}^{n-1}\alpha =-(cot\alpha -\tan \alpha -2\tan 2\alpha -2^2\tan 2^2\alpha -...-2^n\tan 2^2\alpha )+2^{n+1}cot{2}^{n+1}\alpha +cot\alpha =-\left(2cot2\alpha -2\tan 2\alpha -2^2\tan 2^2\alpha -...-2^n\tan 2^2\alpha \right)+2^{n+1}cot{2}^{n+1}\alpha +cot\alpha =-2^{n+1}cot{2}^{n+1}\alpha +2^{n+1}\alpha +cot\alpha =cot\alpha$