If $1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots \ldots=\frac{a-2 \cos \theta}{5+b \cos \theta}$ for some $a, b \in \mathbf{R}$, then $(a-b)^2=$

• A

  0
• Correct answer
  B

  64
• C

  36
• D

  125

Solution

We have, $\begin{aligned} & 1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots=\frac{a-2 \cos \theta}{5+b \cos \theta} \\ & \text { Put, } \theta=0 \\ & \therefore \quad 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \ldots=\frac{a-2}{5+b} \\ & \Rightarrow \quad \frac{1}{1-\frac{1}{2}}=\frac{a-2}{5+b} \Rightarrow 2=\frac{a-2}{5+b} \end{aligned}$
$\begin{aligned} & \text { Put, } \theta=\pi \\ & \begin{aligned} & 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots=\frac{a+2}{5-b} \\ & \Rightarrow \quad \frac{1}{1+\frac{1}{2}}=\frac{a+2}{5-b} \Rightarrow \frac{2}{3}=\frac{a+2}{5-b}\end{aligned}\end{aligned}$
From Eq. (i) and (ii), we get $\begin{aligned} a=4, b & =-4 \\ \therefore \quad(a-b)^2=(4+4)^2 & =8^2=64 \end{aligned}$

Hence, the correct answer is option: 'B'.