Question

The value of $tan\alpha +2tan\left(2\alpha \right)+4tan\left(4\alpha \right)+.....+2^{n-1}tan\left(2^{n-1}\alpha \right)+2^{n}cot\left(2^n\alpha \right)$ is

• A. $cot\left(2^n\alpha \right)$
• B. $2^{n}tan\left(2^n\alpha \right)$
• C. $cot\left(2^n\alpha \right)$
• D. $cot\alpha$

Answer 1

The correct option is D

$cot\alpha$

$cot\alpha -tan\alpha =\frac{co{t}^2\alpha -1}{cot\alpha }=2cot2\alpha tan\alpha =cot\alpha -2cot2\alpha$

Relpacing $\alpha$ by 2$\alpha$ and multiplying both side by 2, we get

$2tan2\alpha =2cot2\alpha -2^{2}cot{2}^2\alpha$

Similiarly,

$2^{2}tan{2}^2\alpha =2^{2}cot{2}^2\alpha -2^{3}cot{2}^3\alpha 2^{n-1}tan{2}^{n-1}\alpha =2^{n-1}cot{2}^{n-1}\alpha -2^{n}cot{2}^n\alpha$

Adding all, we get

$tan\alpha +2tan2\alpha +2^{2}tan{2}^2\alpha +.....2^{n-1}tan{2}^{n-1}\alpha =cot\alpha -2^{n}cot{2}^n\alpha \Rightarrow tan\alpha +2tan2\alpha +2^{2}tan{2}^2\alpha +....+2^{n}cot{2}^n\alpha =cot\alpha$