Q60 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics

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

  • A
    $\sin \alpha$
  • B
    $\cos \alpha$
  • C
    $\tan \alpha$
  • Correct answer
    D
    $\cot \alpha$

Solution

$\begin{aligned} & \text { As, } \tan \theta-\cot \theta=\frac{\sin ^2 \theta-\cos ^2 \theta}{\sin \theta \cos \theta}=-2 \cot 2 \theta \\ & \text { Now, } \tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha \\ & =\tan \alpha-\cot \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha+\cot \alpha \\ & =-2 \cot 2 \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha+\cot \alpha \\ & =-4 \cot 4 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha+\cot \alpha \\ & =-8 \cot 8 \alpha+8 \cot 8 \alpha+\cot \alpha=\cot \alpha\end{aligned}$

Hence, the correct answer is option: 'D'.

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