Question

Let $S_n=\tan \theta +2\tan 2\theta +4\tan 4\theta +\cdots +n\text{terms}$ and $F\left(\theta \right)=\int S_{n}d\theta$. If $\underset{\theta \to 0}{lim}F\left(\theta \right)=1$, then $F\left(\theta \right)$ is

• A. $\ln \left|\frac{2^n\sin \theta }{\sin \left(2^n\theta \right)}\right|$
• B. $\ln \left|\frac{e{2}^n\sin \theta }{\sin \left(2^n\theta \right)}\right|$
• C. $\ln \left|\frac{10\cdot 2^n\sin \theta }{\sin \left(2^n\theta \right)}\right|$
• D. $\ln \left|\frac{\sin \theta }{\sin \left(2^n\theta \right)}\right|$

Answer 1

The correct option is B $\ln \left|\frac{e{2}^n\sin \theta }{\sin \left(2^n\theta \right)}\right|$

$\cot \theta -\tan \theta =\frac{1}{\tan \theta }-\tan \theta =\frac{2(1-{\tan }^2\theta )}{2\tan \theta }=\frac{2}{\tan 2\theta }=2\cot 2\theta$
$\Rightarrow \tan \theta =\cot \theta -2\cot 2\theta$

Replacing $\theta$ by $2^{n-1}\theta$
$\tan \left(2^{n-1}\theta \right)=\cot \left(2^{n-1}\theta \right)-2\cot \left(2^n\theta \right)$

$S_n=\tan \theta +2\tan 2\theta +4\tan 4\theta +\cdots +2^{n-1}\tan \left(2^{n-1}\theta \right)$
$t_n=2^{n-1}\tan \left(2^{n-1}\theta \right)=2^{n-1}\cot \left(2^{n-1}\theta \right)-2^n\cot \left(2^n\theta \right)$

$S_n=t_1+t_2+t_3+\cdots +t_n$
$=(\cot \theta -2\cot 2\theta )+(2\cot 2\theta -4\cot 4\theta )+\cdots +\left[2^{n-1}\cot \right(2^{n-1}\theta )-2^n\cot (2^n\theta \left)\right]$
$=\cot \theta -2^n\cot \left(2^n\theta \right)$

$F\left(\theta \right)=\int S_{n}d\theta =\ln |\sin \theta |-\ln |\sin \left(2^n\theta \right)|+C=\ln \left|\frac{\sin \theta }{\sin \left(2^n\theta \right)}\right|+C$

Now, $\underset{\theta \to 0}{lim}F\left(\theta \right)=1$
$\Rightarrow \underset{\theta \to 0}{lim}\ln \left|\frac{\sin \theta }{\sin \left(2^n\theta \right)}\right|+C=1$

$\Rightarrow \underset{\theta \to 0}{lim}\ln \left|\frac{\frac{\sin \theta }{\theta }\times \theta }{\frac{\sin \left(2^n\theta \right)}{\left(2^n\theta \right)}\times \left(2^n\theta \right)}\right|+C=1$
$\Rightarrow \ln \left(2^{-n}\right)+C=1\Rightarrow -n\ln \left(2\right)+C=1$
$\Rightarrow C=\ln e+n\ln 2$
$\Rightarrow C=\ln \left(e{2}^n\right)$

$\Rightarrow F\left(\theta \right)=\ln \left|\frac{\sin \theta }{\sin \left(2^n\theta \right)}\right|+\ln \left(e{2}^n\right)$
$=\ln \left|\frac{e{2}^n\sin \theta }{\sin \left(2^n\theta \right)}\right|$


Alternate solution:
We know that,
$\int \tan n\theta d\theta =\frac{1}{n}\ln |\sec n\theta |+C\int S_{n}d\theta =\ln |\sec \theta \cdot \sec 2\theta \cdot \sec 4\theta \cdots n\text{terms}|+C$

Now, $\sec \theta \cdot \sec 2\theta \cdot \sec 4\theta \cdots n\text{terms}$

$=\frac{1}{\cos \theta \cdot \cos 2\theta \cdot \cos 4\theta \cdots \cos 2^{n-1}\theta }$

$=\frac{2^n\sin \theta }{\sin \left(2^n\theta \right)}$

So, $\int S_{n}d\theta =\ln \left|\frac{2^n\sin \theta }{\sin \left(2^n\theta \right)}\right|+C$

$\Rightarrow F\left(\theta \right)=\ln \left|\frac{2^n\sin \theta }{\sin \left(2^n\theta \right)}\right|+C$

$\Rightarrow \underset{\theta \to 0}{lim}F\left(\theta \right)=C=1$

$\therefore F\left(\theta \right)=\ln \left|\frac{e{2}^n\sin \theta }{\sin \left(2^n\theta \right)}\right|$