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  4. tan α+2 tan 2 α+4 tan 4 α+8 cot 8 α is equal to :

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Q. $\tan \,\alpha+2\, \tan \,2 \alpha+4 \,\tan \,4 \alpha+8\, \cot \,8 \alpha$ is equal to :

UPSEEUPSEE 2005

Solution:

$\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha$
$=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+\frac{8}{\tan 8 \alpha}$
$=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha$
$+4 \frac{\left(1-\tan ^{2} 4 \alpha\right)}{\tan 4 \alpha}$
$=\tan \alpha+2 \tan 2 \alpha$
$+\left(\frac{4 \tan ^{2} 4 \alpha+4-4 \tan ^{2} 4 \alpha}{\tan 4 \alpha}\right)$
$=\tan \alpha+2 \tan 2 \alpha$
$+\left(\frac{4 \tan ^{2} 4 \alpha+4-4 \tan ^{2} 4 \alpha}{\tan 4 \alpha}\right)$
$=\tan \alpha+2 \tan 2 \alpha+\frac{4}{\tan 4 \alpha}$
$=\tan \alpha+2 \tan 2 \alpha+\frac{2\left(1-\tan ^{2} 2 \alpha\right)}{\tan 2 \alpha}$
$=\tan \alpha+\left(\frac{2 \tan ^{2} 2 \alpha+2-2 \tan ^{2} 2 \alpha}{\tan 2 \alpha}\right)$
$=\tan \alpha+\frac{2}{\tan 2 \alpha}=\tan \alpha+\frac{2\left(1-\tan ^{2} \alpha\right)}{2 \tan \alpha}$
$=\frac{2 \tan ^{2} \alpha+2-2 \tan ^{2} \alpha}{2 \tan \alpha}$
$=\cot \alpha$