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 $\text{cosec}2x+\text{cosec}4x+\text{cosec}8x+\text{cosec}16x+\text{cosec}32x$

• A. $\cot x-\cot 64x$
• B. $\cot 2x-\cot 64x$
• C. $\cot x-\cot 32x$
• D. $\cot 2x-\cot 32x$

Answer 1

The correct option is C $\cot x-\cot 32x$

$\because \text{cosec}2x+\cot 2x=\cot x\Rightarrow \text{cosec}2x=\cot x-\cot 2x$
Now,
$\text{cosec}4x=\cot 2x-\cot 4x\text{cosec}8x=\cot 4x-\cot 8x\text{cosec}16x=\cot 8x-\cot 16x\text{cosec}32x=\cot 16x-\cot 32x$
So,
$\text{cosec}2x+\text{cosec}4x+\text{cosec}8x+\text{cosec}16x+\text{cosec}32x=\cot x-\cot 32x$