Q62 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics
$\cot \theta-\tan \theta-2 \tan 2 \theta-4 \tan 4 \theta=$
- A$4 \cot 8 \theta-\tan 6 \theta$
- B$\cot 8 \theta+\tan 3 \theta$
- C$\cot 86+\cot 6 \theta$
- Correct answerD$8 \cot 86$
Solution
$$
\begin{aligned}
& \cot \theta-\tan \theta-2 \tan 2 \theta-4 \tan 4 \theta \\
& =\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}-2 \frac{\sin 2 \theta}{\cos 2 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta} \\
& =2 \frac{\cos ^2 \theta-\sin ^2 \theta}{2 \sin \theta \cos \theta}-2 \frac{\sin 2 \theta}{\cos 2 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta} \\
& =2 \frac{\cos 2 \theta}{\sin 2 \theta}-\frac{2 \sin 2 \theta}{\cos 2 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad 2 \cdot 2 \frac{\left[\cos ^2 2 \theta-\sin ^2 2 \theta\right]}{2 \sin 2 \theta \cdot \cos 2 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta} \\
& \Rightarrow \frac{4 \cos 4 \theta}{\sin 4 \theta}-4 \frac{\sin 4 \theta}{\cos 4 \theta}=4\left[\frac{\cos ^2 4 \theta-\sin ^2 4 \theta}{\sin 4 \theta \cos 4 \theta}\right]=8 \frac{\cos 8 \theta}{\sin 8 \theta} \\
& \Rightarrow \quad 8 \cot 8 \theta
\end{aligned}
$$
Hence, the correct answer is option: 'D'.