$\tan A+2 \tan 2 A+4 \tan 4 A+8 \cot 8 A=$

$\tan A+2 \tan 2 A+4 \tan 4 A+8 \cot 8 A=$
  1. $\tan 2 \mathrm{~A}$
  2. $\cot \mathrm{A}$
  3. $\tan \mathrm{A}$
  4. $\cot 2 \mathrm{~A}$

Solution

$\begin{aligned} & \tan A+2 \tan 2 A+4 \tan 4 A+8 \cot 8 A \\ & =\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+4 \tan 4 \mathrm{~A}+\frac{8}{\left(\frac{2 \tan 4 \mathrm{~A}}{1-\tan ^2 4 \mathrm{~A}}\right)} \\ & =\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+4 \tan 4 \mathrm{~A}+\frac{8\left(1-\tan ^2 \mathrm{~A}\right)}{2 \tan 4 \mathrm{~A}} \\ & =\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+\frac{(4 \tan 4 \mathrm{~A})(2 \tan 4 \mathrm{~A})+8\left(1-\tan ^2 4 \mathrm{~A}\right)}{2 \tan 4 \mathrm{~A}} \\ & =\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+\frac{8}{2 \tan 4 \mathrm{~A}} \\ & =\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+\frac{8}{2\left(\frac{2 \tan \mathrm{A}}{1-\tan ^2 2 \mathrm{~A}}\right)} \\ & =\tan \mathrm{A}+2 \tan 2 \mathrm{~A}+\frac{8\left(1-\tan ^2 2 \mathrm{~A}\right)}{4 \tan 2 \mathrm{~A}} \\ & =\tan \mathrm{A}+\frac{(2 \tan 2 \mathrm{~A})(4 \tan 2 \mathrm{~A})+8\left(1-\tan ^2 2 \mathrm{~A}\right)}{4 \tan 2 \mathrm{~A}}=\tan \mathrm{A}+\frac{8}{4 \tan 2 \mathrm{~A}} \\ & =\tan \mathrm{A}+\frac{8}{4\left(\frac{2 \tan \mathrm{A}}{1-\tan ^2 \mathrm{~A}}\right)}=\tan \mathrm{A}+\frac{8\left(1-\tan ^2 \mathrm{~A}\right)}{8 \tan \mathrm{A}} \\ & \end{aligned}$ $=\frac{\tan \mathrm{A}(8 \tan \mathrm{A})+8\left(1-\tan ^2 \mathrm{~A}\right)}{8 \tan \mathrm{A}}=\frac{8}{8 \tan \mathrm{A}}=\cot \mathrm{A}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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