Solve the following

Solution:

We need to prove $\frac{1}{2} \tan \left(\frac{x}{2}\right)+\frac{1}{4} \tan \left(\frac{x}{4}\right)+\ldots+\frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{1}{2^{n}} \cot \left(\frac{x}{2^{n}}\right)-\cot x$ for all $n \in N$ and $0

For $n=1$,

$\mathrm{LHS}=\frac{1}{2} \tan \frac{x}{2}$

And,

$\mathrm{RHS}=\frac{1}{2} \cot \frac{x}{2}-\cot x=\frac{1}{2 \tan \frac{x}{2}}-\frac{1}{\tan x}$

$\Rightarrow \mathrm{RHS}=\frac{1}{2 \tan \frac{x}{2}}-\frac{1}{\frac{2 \tan \frac{x}{2}}{1-\tan ^{2} \frac{x}{2}}}$

$\Rightarrow \mathrm{RHS}=\frac{1}{2 \tan \frac{x}{2}}-\frac{1-\tan ^{2} \frac{x}{2}}{2 \tan \frac{x}{2}}=\frac{1-1+\tan ^{2} \frac{x}{2}}{2 \tan \frac{x}{2}}=\frac{\tan \frac{x}{2}}{2}$

Therefore, the given relation is true for $n=1$.

Now, let the given relation be true for $n=k$.

We need to prove that the given relation is true for $n=k+1$.

$\therefore \frac{1}{2} \tan \left(\frac{x}{2}\right)+\frac{1}{4} \tan \left(\frac{x}{4}\right)+\ldots+\frac{1}{2^{k}} \tan \left(\frac{x}{2^{k}}\right)=\frac{1}{2^{k}} \cot \left(\frac{x}{2^{k}}\right)-\cot x$

Now,

$\frac{1}{2} \tan \left(\frac{x}{2}\right)+\frac{1}{4} \tan \left(\frac{x}{4}\right)+\ldots+\frac{1}{2^{k}} \tan \left(\frac{x}{2^{k}}\right)+\frac{1}{2^{k+1}} \tan \left(\frac{x}{2^{k+1}}\right)=\frac{1}{2^{k}} \cot \left(\frac{x}{2^{k}}\right)-\cot x+\frac{1}{2^{k+1}} \tan \left(\frac{x}{2^{k+1}}\right)$

Let:

$L=\frac{1}{2^{k}} \cot \left(\frac{x}{2^{k}}\right)-\cot x+\frac{1}{2^{k+1}} \tan \left(\frac{x}{2^{k+1}}\right)$

$\Rightarrow L=\frac{1}{2^{k}} \cot \left(\frac{x}{2^{k}}\right)+\frac{1}{2^{k+1}} \tan \left(\frac{x}{2^{k+1}}\right)-\cot x$

$\Rightarrow L=\frac{1}{2^{k} \tan \left(\frac{x}{2^{k}}\right)}+\frac{1}{2^{k+1}} \tan \left(\frac{x}{2^{k+1}}\right)-\cot x$

$\Rightarrow L=\frac{1}{2^{k} \tan 2\left(\frac{x}{2^{k+1}}\right)}+\frac{1}{2^{k+1}} \tan \left(\frac{x}{2^{k+1}}\right)-\cot x$

$\Rightarrow L=\frac{1}{2^{k} \times \frac{2 \tan \left(\frac{x}{2^{k+1}}\right)}{1-\tan ^{2}\left(\frac{x}{2^{k+1}}\right)}}+\frac{1}{2^{k+1}} \tan \left(\frac{x}{2^{k+1}}\right)-\cot x$

$\Rightarrow L=\frac{1-\tan ^{2}\left(\frac{x}{2^{k+1}}\right)}{2^{k+1} \tan \left(\frac{x}{2^{k+1}}\right)}+\frac{1}{2^{k+1}} \tan \left(\frac{x}{2^{k+1}}\right)-\cot x$

$\Rightarrow L=\frac{1-\tan ^{2}\left(\frac{x}{2^{k+1}}\right)+\tan ^{2}\left(\frac{x}{2^{k+1}}\right)}{2^{k+1} \tan \left(\frac{x}{2^{k+1}}\right)}-\cot x=\frac{1}{2^{k+1}} \cot \left(\frac{x}{2^{k+1}}\right)-\cot x$

Now,

$\frac{1}{2} \tan \left(\frac{x}{2}\right)+\frac{1}{4} \tan \left(\frac{x}{4}\right)+\ldots+\frac{1}{2^{k}} \tan \left(\frac{x}{2^{k}}\right)+\frac{1}{2^{k+1}} \tan \left(\frac{x}{2^{k+1}}\right)=\frac{1}{2^{k+1}} \cot \left(\frac{x}{2^{k+1}}\right)-\cot x$

Thus, $\frac{1}{2} \tan \left(\frac{x}{2}\right)+\frac{1}{4} \tan \left(\frac{x}{4}\right)+\ldots+\frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{1}{2^{n}} \cot \left(\frac{x}{2^{n}}\right)-\cot x$ for all $n \in N$ and $0