Q61 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=$
- Correct answerA$4$
- B$3$
- C$2$
- D$1$
Solution
$\begin{aligned} & \tan 81^{\circ}+\tan 9^{\circ}-\tan 63^{\circ}-\tan 27^{\circ} \\ = & \cot 9^{\circ}+\tan 9^{\circ}-\left(\cot 27^{\circ}+\tan 27^{\circ}\right) \\ = & \frac{\left(\cos 9^{\circ}\right)^2+\left(\sin 9^{\circ}\right)^2}{\sin 9^{\circ} \cdot \cos 9^{\circ}}-\frac{\left(\cos 27^{\circ}\right)^2+\left(\sin 27^{\circ}\right)^2}{\sin 27^{\circ} \cdot \cos 27^{\circ}} \\ = & \frac{1}{\sin 9^{\circ} \cdot \cos 9^{\circ}}-\frac{1}{\sin 27^{\circ} \cos 27^{\circ}}=\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}} \\ = & \frac{2\left(\sin 54^{\circ}-\sin 18^{\circ}\right)}{\sin 54^{\circ} \sin 18^{\circ}}=\frac{4 \cos 36^{\circ} \sin 18^{\circ}}{\cos 36^{\circ} \sin 18^{\circ}}=4\end{aligned}$
Hence, the correct answer is option: 'A'.