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AP EAPCET 19-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 21 of 160

Marks: +1, -0

\tan \alpha+2 \tan 2\alpha+4 \tan 4\alpha+8 \cot 8\alpha=

$\tan 16\alpha$
0
$\cot \alpha$
$\tan \alpha$

Solution:

Let

\tan A+2 \cot 2A

=\frac{\sin A}{\cos A}+\frac{2 \cos 2A}{\sin 2A}

=\frac{\sin A}{\cos A}+\frac{2(1-2 \sin^2 A)}{2 \sin A \cos A}

=\frac{\sin^2 A+1-2 \sin^2 A}{\sin A \cos A}

=\frac{1-\sin^2 A}{\sin A \cos A}=\frac{\cos^2 A}{\sin A \cos A}=\cot A

∴

\tan \alpha+2 \tan 2\alpha+4 \tan 4\alpha+8 \cot 8\alpha

=\tan \alpha+2 \tan 2\alpha+4(\tan 4\alpha+2 \cot 2(4\alpha))

=\tan \alpha+2 \tan 2\alpha+4 \cot 4\alpha

=\tan \alpha+2(\tan 2\alpha+2 \cot 2(2\alpha))

=\tan \alpha+2 \cot 2\alpha=\cot \alpha

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