Question

$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8cot8\alpha =$

• A. $\tan \alpha$
• B. $\tan 2\alpha$
• C. $cot\alpha$
• D. $cot2\alpha$

Answer 1

The correct option is C

$cot\alpha$

Explanation for the correct answer:

Simplifying the given expression:

$$\begin{gathered} \tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\mathrm{\alpha } \\ =\tan \mathrm{\alpha }+2\tan 2\mathrm{\alpha }+4\tan 4\mathrm{\alpha }+8\left(\frac{1}{\tan 8\mathrm{\alpha }}\right)\left[\because \cot x=\frac{1}{\tan x}\right] \\ =\tan \mathrm{\alpha }+2\tan 2\mathrm{\alpha }+4\tan 4\mathrm{\alpha }+8\left\{\frac{1}{\tan 2\left(4\mathrm{\alpha }\right)}\right\} \\ =\tan \mathrm{\alpha }+2\tan 2\mathrm{\alpha }+4\tan 4\mathrm{\alpha }+8\left[\frac{1}{{\displaystyle \frac{2\tan \left(4\mathrm{\alpha }\right)}{1-{\tan }^2\left(4\mathrm{\alpha }\right)}}}\right]\left[\because \tan 2\mathrm{A}=\frac{2\tan \mathrm{A}}{(1-{\tan }^2\mathrm{A})}\right] \\ =\tan \mathrm{\alpha }+2\tan 2\mathrm{\alpha }+4\tan 4\mathrm{\alpha }+8\left[\frac{\left\{1-{\tan }^2\left(4\mathrm{\alpha }\right)\right\}}{2\tan 4\mathrm{\alpha }}\right] \\ =\tan \mathrm{\alpha }+2\tan 2\mathrm{\alpha }+\frac{\left[4{\tan }^2\left(4\mathrm{\alpha }\right)+4-4{\tan }^2\left(4\mathrm{\alpha }\right)\right]}{\tan 4\mathrm{\alpha }} \\ =\tan \mathrm{\alpha }+2\tan 2\mathrm{\alpha }+4\left(\frac{1}{\tan 4\mathrm{\alpha }}\right) \\ =\tan \mathrm{\alpha }+2\tan 2\mathrm{\alpha }+4\left[\frac{\left\{1-{\tan }^2(2\mathrm{\alpha }\right\}}{2\tan 2\mathrm{\alpha }}\right]\left[\because \tan 2\mathrm{A}=\frac{2\tan \mathrm{A}}{(1-{\tan }^2\mathrm{A})}\right] \\ =\tan \mathrm{\alpha }+\frac{\left[2{\tan }^2\left(2\mathrm{\alpha }\right)+2-2{\tan }^2(2\mathrm{\alpha })\right]}{\tan 2\mathrm{\alpha }} \\ =\tan \mathrm{\alpha }+\frac{2}{\tan 2\mathrm{\alpha }} \\ =\tan \mathrm{\alpha }+2\left[\frac{(1-{\tan }^2\mathrm{\alpha })}{2\tan \mathrm{\alpha }}\right]\left[\because \tan 2\mathrm{A}=\frac{2\tan \mathrm{A}}{(1-{\tan }^2\mathrm{A})}\right] \\ =\frac{\left[{\tan }^2\mathrm{\alpha }+1-{\tan }^2\mathrm{\alpha }\right]}{\tan \mathrm{\alpha }} \\ =\frac{1}{\tan \mathrm{\alpha }} \\ =\cot \mathrm{\alpha }\left[\because \cot x=\frac{1}{\tan x}\right] \end{gathered}$$

Therefore, the correct answer is option (C).