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Question: The value of the expression \(\tan \alpha +2\tan \left( 2\alpha \right)+4\tan \left( 4\alpha \right)...

The value of the expression tan⁡α+2tan⁡(2α)+4tan⁡(4α)+...+2n−1tan⁡(2n−1α)+2ncot⁡(2nα)\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⁡(2nα)\cot \left( {{2}^{n}}\alpha \right)
  2. 2ntan⁡(2nα){{2}^{n}}\tan \left( {{2}^{n}}\alpha \right)
  3. 00
  4. cot⁡α\cot \alpha
Explanation

Solution

For finding the value of the given question tan⁡α+2tan⁡(2α)+4tan⁡(4α)+...+2n−1tan⁡(2n−1α)+2ncot⁡(2nα)\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⁡α\tan \alpha , 2tan⁡2α2\tan 2\alpha , ….2n−1tan⁡(2n−1α){{2}^{n-1}}\tan \left( {{2}^{n-1}}\alpha \right) with the use of cot⁡α−tan⁡α\cot \alpha -\tan \alpha . Similarly, we will get the value of 2tan⁡2α2\tan 2\alpha , ….2n−1tan⁡(2n−1α){{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:
⇒tan⁡α+2tan⁡(2α)+4tan⁡(4α)+...+2n−1tan⁡(2n−1α)+2ncot⁡(2nα)\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⁡α\tan \alpha with the help of getting sum of tan⁡α+cot⁡α\tan \alpha +\cot \alpha as:
⇒cot⁡α−tan⁡α\Rightarrow \cot \alpha -\tan \alpha
As we know that tan⁡α\tan \alpha is equal to 1cot⁡α\dfrac{1}{\cot \alpha } . So, we can write the above step below as:
⇒cot⁡α−1cot⁡α\Rightarrow \cot \alpha -\dfrac{1}{\cot \alpha }
Now, we take L.C.M. from denominator to solve the above step as:
⇒cot⁡2α−1cot⁡α\Rightarrow \dfrac{{{\cot }^{2}}\alpha -1}{\cot \alpha }
By using the formula of cot⁡2α\cot 2\alpha , we can write the above step below as:
⇒2cot⁡2α\Rightarrow 2\cot 2\alpha
Now, we can write the whole equation as:
⇒cot⁡α−tan⁡α=2cot⁡2α\Rightarrow \cot \alpha -\tan \alpha =2\cot 2\alpha
From the above equation, we will get the value of tan⁡α\tan \alpha as:
⇒tan⁡α=cot⁡α−2cot⁡2α\Rightarrow \tan \alpha =\cot \alpha -2\cot 2\alpha … (1)\left( 1 \right)
Now, similarly we can calculate the value of tan⁡2α\tan 2\alpha and will have the equation as:
⇒tan⁡2α=cot⁡2α−2cot⁡4α\Rightarrow \tan 2\alpha =\cot 2\alpha -2\cot 4\alpha
Here, we will multiply by 22 both sides of the above equation and will get:
⇒2tan⁡2α=2cot⁡2α−4cot⁡4α\Rightarrow 2\tan 2\alpha =2\cot 2\alpha -4\cot 4\alpha … (2)\left( 2 \right)
Similarly, if we replace α\alpha from 4α4\alpha and multiply by 44 in equation (1)\left( 1 \right) will have:
⇒4tan⁡4α=4cot⁡4α−8cot⁡8α\Rightarrow 4\tan 4\alpha =4\cot 4\alpha -8\cot 8\alpha … (3)\left( 3 \right)
And we will continue this process up to (n−1)α\left( n-1 \right)\alpha and will multiply by (n−1)\left( n-1 \right) in equation (1)\left( 1 \right) and will get the equation as:
⇒2(n−1)tan⁡(2(n−1)α)=2(n−1)cot⁡(2(n−1)α)−2(n−1+1)cot⁡(2(n−1+1)α)\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:
⇒2(n−1)tan⁡(2(n−1)α)=2(n−1)cot⁡(2(n−1)α)−2ncot⁡(2nα)\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) … (n−1)\left( n-1 \right)
Now, we will add both sides of all the equations from (1)\left( 1 \right) to (n−1)\left( n-1 \right) and will get 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\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{aligned}$$ After simplifying the above equation in which we will cancel out some terms, we will have the above equation as: $$\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{aligned} & \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{aligned}$$ In the above step, we got the expansion of the formula $\cos \left( {{2}^{n}}-{{2}^{n-1}} \right)\alpha $. So, we will use it as: $$\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.