Question

For all permissible values of $A,2A$, following holds true.
$\left(i\right)\cot A+\tan A=\frac{1}{\sin A\cos A}=2\text{cosec}2A\left(ii\right)\cot A-\tan A=\frac{{\cos }^{2}A-{\sin }^{2}A}{\sin A\cos A}=2\cot 2A\left(iii\right)2\cot A=2(\text{cosec}2A+\cot 2A)\Rightarrow \text{cosec}2A+\cot 2A=\cot A$

Also to evaluate a series of form $f\left(x\right)+f\left(2x\right)+f\left(4x\right)+\cdots +f\left(2^{n}x\right)$ when $f\left(x\right)$ can be expressed as $g\left(x\right)-g\left(2x\right)$, we can use the following technique,
$f\left(x\right)+f\left(2x\right)+f\left(4x\right)+\cdots +f\left(2^{n}x\right)=\left(g\right(x)-g(2x\left)\right)+\left(g\right(2x)-g(4x\left)\right)+\cdots \left(g\right(2^{n}x)-g(2^{n+1}x\left)\right)=g\left(x\right)-g\left(2^{n+1}x\right)$

Based on the above information, solve the following questions for all permissible values of $x$.

The value of $\tan x+\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\cdots +\frac{1}{2^{10}}\tan \frac{x}{2^{10}}$ is

• A. $\frac{1}{2^{11}}\cot \frac{x}{2^{11}}-2\cot 2x$
• B. $\frac{1}{2^{10}}\cot \frac{x}{2^{10}}-2\cot 2x$
• C. $\frac{1}{2^{10}}\cot \frac{x}{2^{10}}-2\cot x$
• D. None of the above

Answer 1

The correct option is B $\frac{1}{2^{10}}\cot \frac{x}{2^{10}}-2\cot 2x$

$\because \cot A-\tan A=2\cot 2A\Rightarrow \tan A=\cot A-2\cot 2A$
Now,
$\tan x=\cot x-2\cot 2x\frac{1}{2}\tan \frac{x}{2}=\frac{1}{2}(\cot \frac{x}{2}-2\cot x)\frac{1}{2^2}\tan \frac{x}{2^2}=\frac{1}{2^2}(\cot \frac{x}{2^2}-2\cot \frac{x}{2})...\frac{1}{2^{10}}\tan \frac{x}{2^{10}}=\frac{1}{2^{10}}(\cot \frac{x}{2^{10}}-2\cot \frac{x}{2^9})$
So,
$\tan x+\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\cdots +\frac{1}{2^{10}}\tan \frac{x}{2^{10}}=\frac{1}{2^{10}}\cot \frac{x}{2^{10}}-2\cot 2x$