Q63 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
$\sin 20^{\circ}\left(4+\sec 20^{\circ}\right)=$
- Correct answerA$\sqrt{3}$
- B$-\sqrt{3}$
- C1
- D-1
Solution
$\sin 20^{\circ}\left[4+\sec 20^{\circ}\right]$
$\begin{aligned} & =\sin 20^{\circ}\left[\frac{4 \cos 20^{\circ}+1}{\cos 20^{\circ}}\right]=\frac{4 \sin 20^{\circ} \cos 20^{\circ}+\sin 20^{\circ}}{\cos 20^{\circ}} \\ & =\frac{2 \sin 40^{\circ}+\sin 20^{\circ}}{\cos 20^{\circ}}=\frac{2 \sin (60-20)^{\circ}+\sin 20^{\circ}}{\cos 20^{\circ}} \\ & =\frac{2\left[\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ}\right]+\sin 20^{\circ}}{\cos 20^{\circ}} \\ & =\frac{2\left[\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}\right]+\sin 20^{\circ}}{\cos 20^{\circ}}=\frac{\sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ}}=\sqrt{3}\end{aligned}$
Hence, the correct answer is option: 'A'.