Question

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

• A. $\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)$
• B. $2^n\tan \left(2^{\mathrm{n}}\mathrm{\alpha }\right)$
• C. $0$
• D. $\cot \left(\mathrm{\alpha }\right)$

Answer 1

The correct option is D

$\cot \left(\mathrm{\alpha }\right)$

The explanation for the correct option:

Compute the required value:

The given trigonometric expression: $\tan \left(\alpha \right)+2\tan \left(2\alpha \right)+4\tan \left(4\alpha \right)+...+2^{n-1}\tan \left(2^{n-1}\alpha \right)+2^n\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)$.

Now, $\cot \left(A\right)-\tan \left(A\right)=\cot \left(A\right)-\frac{1}{\cot \left(A\right)}\left[\because \tan \left(\theta \right)=\frac{1}{\cot \left(\theta \right)}\right]$

$$\begin{gathered} \Rightarrow \cot \left(A\right)-\tan \left(A\right)=\frac{{\cot }^2\left(A\right)-1}{\cot \left(A\right)} \\ \Rightarrow \cot \left(A\right)-\tan \left(A\right)=2\times \frac{{\cot }^2\left(A\right)-1}{2\cot \left(A\right)} \\ \Rightarrow \cot \left(A\right)-\tan \left(A\right)=2\cot \left(2A\right)\left[\because \cot \left(2\theta \right)=\frac{{\cot }^2\left(\theta \right)-1}{2\cot \left(\theta \right)}\right] \end{gathered}$$

Thus, $\tan \left(\alpha \right)+2\tan \left(2\alpha \right)+4\tan \left(4\alpha \right)+...+2^{n-1}\tan \left(2^{n-1}\alpha \right)+2^n\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)=\tan \left(\alpha \right)-\cot \left(\mathrm{\alpha }\right)+2\tan \left(2\alpha \right)+4\tan \left(4\alpha \right)+...+2^{n-1}\tan \left(2^{n-1}\alpha \right)+2^n\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)+\cot \left(\mathrm{\alpha }\right)$

$$\begin{gathered} \Rightarrow \tan \left(\alpha \right)+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^{\mathrm{n}}\mathrm{\alpha }\right)=-2\cot \left(2\mathrm{\alpha }\right)+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^{\mathrm{n}}\mathrm{\alpha }\right)+\cot \left(\mathrm{\alpha }\right) \\ \Rightarrow \tan \left(\alpha \right)+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^{\mathrm{n}}\mathrm{\alpha }\right)=-4\cot \left(4\mathrm{\alpha }\right)+4\tan \left(4\alpha \right)+...+2^{n-1}\tan \left(2^{n-1}\alpha \right)+2^n\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)+\cot \left(\mathrm{\alpha }\right) \\ \Rightarrow \tan \left(\alpha \right)+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^{\mathrm{n}}\mathrm{\alpha }\right)=-8\cot \left(8\mathrm{\alpha }\right)+...+2^{n-1}\tan \left(2^{n-1}\alpha \right)+2^n\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)+\cot \left(\mathrm{\alpha }\right) \\ \Rightarrow \tan \left(\alpha \right)+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^{\mathrm{n}}\mathrm{\alpha }\right)=-2^{n-1}\cot \left(2^{\mathrm{n}-1}\mathrm{\alpha }\right)+2^{n-1}\tan \left(2^{n-1}\alpha \right)+2^n\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)+\cot \left(\mathrm{\alpha }\right) \\ \Rightarrow \tan \left(\alpha \right)+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^{\mathrm{n}}\mathrm{\alpha }\right)=-2^n\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)+2^n\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)+\cot \left(\mathrm{\alpha }\right) \\ \Rightarrow \tan \left(\alpha \right)+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^{\mathrm{n}}\mathrm{\alpha }\right)=\cot \left(\mathrm{\alpha }\right) \end{gathered}$$

Hence, the value of $\tan \left(\alpha \right)+2\tan \left(2\alpha \right)+4\tan \left(4\alpha \right)+...+2^{n-1}\tan \left(2^{n-1}\alpha \right)+2^n\cot \left(2^{\mathrm{n}}\mathrm{\alpha }\right)$ is $\cot \left(\mathrm{\alpha }\right)$.

Hence, option $D$ is the correct answer.