Question

The value of the expression $\tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right)+{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right)$ is
1) $\cot \left( {{2}^{n}}\alpha \right)$
2) ${{2}^{n}}\tan \left( {{2}^{n}}\alpha \right)$
3) $0$
4) $\cot \alpha $

Answer 1

Hint: For finding the value of the given question $\tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right)+{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right)$ , we will use a different method. In which we will try to find the value of $\tan \alpha $ , $2\tan 2\alpha $, ….${{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right)$ with the use of $\cot \alpha -\tan \alpha $ . Similarly, we will get the value of $2\tan 2\alpha $, ….${{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right)$ and so on. After adding them and then simplifying, we will get the value of the given equation.

Complete step-by-step solution:
Since, the given question, we have:
$\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right)+{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right)$
Here, we need to have to find the value of this question. So, we will try to find the value of $\tan \alpha $ with the help of getting sum of $\tan \alpha +\cot \alpha $ as:
$\Rightarrow \cot \alpha -\tan \alpha $
As we know that $\tan \alpha $ is equal to $\dfrac{1}{\cot \alpha }$ . So, we can write the above step below as:
$\Rightarrow \cot \alpha -\dfrac{1}{\cot \alpha }$
Now, we take L.C.M. from denominator to solve the above step as:
$\Rightarrow \dfrac{{{\cot }^{2}}\alpha -1}{\cot \alpha }$
By using the formula of $\cot 2\alpha $ , we can write the above step below as:
$\Rightarrow 2\cot 2\alpha $
Now, we can write the whole equation as:
$\Rightarrow \cot \alpha -\tan \alpha =2\cot 2\alpha $
From the above equation, we will get the value of $\tan \alpha $ as:
$\Rightarrow \tan \alpha =\cot \alpha -2\cot 2\alpha $ … $\left( 1 \right)$
Now, similarly we can calculate the value of $\tan 2\alpha $ and will have the equation as:
$\Rightarrow \tan 2\alpha =\cot 2\alpha -2\cot 4\alpha $
Here, we will multiply by $2$ both sides of the above equation and will get:
$\Rightarrow 2\tan 2\alpha =2\cot 2\alpha -4\cot 4\alpha $ … $\left( 2 \right)$
Similarly, if we replace $\alpha $ from $4\alpha $ and multiply by $4$ in equation $\left( 1 \right)$ will have:
$\Rightarrow 4\tan 4\alpha =4\cot 4\alpha -8\cot 8\alpha $ … $\left( 3 \right)$
And we will continue this process up to $\left( n-1 \right)\alpha $ and will multiply by $\left( n-1 \right)$ in equation $\left( 1 \right)$ and will get the equation as:
$\Rightarrow {{2}^{\left( n-1 \right)}}\tan \left( {{2}^{\left( n-1 \right)}}\alpha \right)={{2}^{\left( n-1 \right)}}\cot \left( {{2}^{\left( n-1 \right)}}\alpha \right)-{{2}^{\left( n-1+1 \right)}}\cot \left( {{2}^{\left( n-1+1 \right)}}\alpha \right)$
After simplifying the power, we will have the above equation as:
$\Rightarrow {{2}^{\left( n-1 \right)}}\tan \left( {{2}^{\left( n-1 \right)}}\alpha \right)={{2}^{\left( n-1 \right)}}\cot \left( {{2}^{\left( n-1 \right)}}\alpha \right)-{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right)$ … $\left( n-1 \right)$
Now, we will add both sides of all the equations from $\left( 1 \right)$ to $\left( n-1 \right)$ and will get the above equation as:
\[\begin{align}
  & \Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right) \\
 & =\cot \alpha -2\cot 2\alpha +2\cot 2\alpha -4\cot 4\alpha +4\cot 4\alpha -8\cot 8\alpha +{{2}^{\left( n-1 \right)}}\cot \left( {{2}^{\left( n-1 \right)}}\alpha \right)-{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right) \\
\end{align}\]
After simplifying the above equation in which we will cancel out some terms, we will have the above equation as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right)=\cot \alpha -{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right)\]
Here, we will change the place of term \[{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right)\] from right side to left side and the equation will be as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right)+{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right)=\cot \alpha \]
Hence, this is the solution.

Note: Here, we will solve the given equation in the other way as:
Since, the question is:
$\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right)+{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right)$
Now, we will simplify the last two terms as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+\left[ {{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right)+{{2}^{n}}\cot \left( {{2}^{n}}\alpha \right) \right]\]
Since, we know that we can write $\tan \alpha $ and $\cot \alpha $ in the form of $\sin \alpha $ and $\cos \alpha $ as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+\left[ {{2}^{n-1}}\dfrac{\sin \left( {{2}^{n-1}}\alpha \right)}{\cos \left( {{2}^{n-1}}\alpha \right)}+{{2}^{n}}\dfrac{\cos \left( {{2}^{n}}\alpha \right)}{\sin \left( {{2}^{n}}\alpha \right)} \right]\]
Further we will simplify as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\left[ \dfrac{\sin \left( {{2}^{n-1}}\alpha \right)\sin \left( {{2}^{n}}\alpha \right)+2\cos \left( {{2}^{n-1}}\alpha \right)\cos \left( {{2}^{n}}\alpha \right)}{\cos \left( {{2}^{n-1}}\alpha \right)\sin \left( {{2}^{n}}\alpha \right)} \right]\]
\[\begin{align}
  & \Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+... \\
 & +{{2}^{n-1}}\left[ \dfrac{\sin \left( {{2}^{n-1}}\alpha \right)\sin \left( {{2}^{n}}\alpha \right)+\cos \left( {{2}^{n-1}}\alpha \right)\cos \left( {{2}^{n}}\alpha \right)+\cos \left( {{2}^{n-1}}\alpha \right)\cos \left( {{2}^{n}}\alpha \right)}{\cos \left( {{2}^{n-1}}\alpha \right)\sin \left( {{2}^{n}}\alpha \right)} \right] \\
\end{align}\]
In the above step, we got the expansion of the formula $\cos \left( {{2}^{n}}-{{2}^{n-1}} \right)\alpha $. So, we will use it as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\left[ \dfrac{\cos \left( {{2}^{n}}-{{2}^{n-1}} \right)\alpha +\cos \left( {{2}^{n-1}}\alpha \right)\cos \left( {{2}^{n}}\alpha \right)}{\cos \left( {{2}^{n-1}}\alpha \right)\sin \left( {{2}^{n}}\alpha \right)} \right]\]
Now, we take ${{2}^{n-1}}$ common as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\left[ \dfrac{\cos {{2}^{n-1}}\left( 2-1 \right)\alpha +\cos \left( {{2}^{n-1}}\alpha \right)\cos \left( {{2}^{n}}\alpha \right)}{\cos \left( {{2}^{n-1}}\alpha \right)\sin \left( {{2}^{n}}\alpha \right)} \right]\]
After solving it, we will have:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\left[ \dfrac{\cos {{2}^{n-1}}\alpha +\cos \left( {{2}^{n-1}}\alpha \right)\cos \left( {{2}^{n}}\alpha \right)}{\cos \left( {{2}^{n-1}}\alpha \right)\sin \left( {{2}^{n}}\alpha \right)} \right]\]
Now, we will take common \[\cos {{2}^{n-1}}\alpha \] in the numerator and will write the above step as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\left[ \dfrac{\cos \left( {{2}^{n-1}}\alpha \right)\left[ 1+\cos \left( {{2}^{n}}\alpha \right) \right]}{\cos \left( {{2}^{n-1}}\alpha \right)\sin \left( {{2}^{n}}\alpha \right)} \right]\]
From the numerator and denominator, \[\cos {{2}^{n-1}}\alpha \] will be canceling out as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\left[ \dfrac{\left[ 1+\cos \left( {{2}^{n}}\alpha \right) \right]}{\sin \left( {{2}^{n}}\alpha \right)} \right]\]
We can write \[\cos \left( {{2}^{n}}\alpha \right)\] as \[2{{\cos }^{2}}\left( {{2}^{n-1}} \right)\alpha -1\] and \[\sin \left( {{2}^{n}}\alpha \right)\] as $2\sin {{\left( 2 \right)}^{n-1}}\alpha .\cos {{\left( 2 \right)}^{n-1}}\alpha $ . Thus, above equation will be as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\left[ \dfrac{\left[ 1+2{{\cos }^{2}}\left( {{2}^{n-1}} \right)\alpha -1 \right]}{2\sin {{\left( 2 \right)}^{n-1}}\alpha .\cos {{\left( 2 \right)}^{n-1}}\alpha } \right]\]
After solving it we will have the above equation as:
\[\Rightarrow \tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)+...+{{2}^{n-1}}\cot \left( {{2}^{n-1}} \right)\alpha \]
Similarly, we will combine all the term up to \[2\tan \left( 2\alpha \right)\] and will get in last the equation as:
\[\Rightarrow \tan \alpha +2\cot 2\alpha \]
And similarly, we will solve it as we did for last two terms and will have the value as:
\[\Rightarrow \cot \alpha \]
Hence, the solution is correct.