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 $\cot 37{\frac{1}{2}}^{\circ }$ is

• A. $\sqrt{6}+\sqrt{4}+\sqrt{3}-\sqrt{2}$
• B. $\sqrt{6}+\sqrt{4}-\sqrt{3}-\sqrt{2}$
• C. $\sqrt{6}+\sqrt{4}+\sqrt{3}+\sqrt{2}$
• D. $\sqrt{6}-\sqrt{4}-\sqrt{3}-\sqrt{2}$

Answer 1

The correct option is B $\sqrt{6}+\sqrt{4}-\sqrt{3}-\sqrt{2}$

$\cot A=\text{cosec}2A+\cot 2A$
Putting $A=37{\frac{1}{2}}^{\circ }$,
$\cot 37{\frac{1}{2}}^{\circ }=\text{cosec}{75}^{\circ }+\cot {75}^{\circ }\Rightarrow \cot 37{\frac{1}{2}}^{\circ }=\frac{2\sqrt{2}}{\sqrt{3}+1}+2-\sqrt{3}\Rightarrow \cot 37{\frac{1}{2}}^{\circ }=\sqrt{6}-\sqrt{2}+2-\sqrt{3}\Rightarrow \cot 37{\frac{1}{2}}^{\circ }=\sqrt{6}+\sqrt{4}-\sqrt{2}-\sqrt{3}$