Q33 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
The value of $\frac{\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}}$ is equal to
- A16
- B32
- Correct answerC64
- D12
Solution
Numerator:
$\sqrt{3}\operatorname{cosec} 20° - \sec 20° = \frac{\sqrt{3}\cos 20° - \sin 20°}{\sin 20° \cos 20°}$
$\sqrt{3}\cos 20° - \sin 20° = 2\left(\frac{\sqrt{3}}{2}\cos 20° - \frac{1}{2}\sin 20°\right)$
$= 2\cos(20° + 30°) = 2\cos 50°$
So numerator = $\frac{2\cos 50°}{\sin 20° \cos 20°} = \frac{4\cos 50°}{\sin 40°} = 4$
Denominator:
$\cos 20° \cos 40° \cos 60° \cos 80°$ where $\cos 60° = \frac{1}{2}$
Using product formula:
$\sin 20° \cos 20° \cos 40° \cos 80° = \frac{1}{8}\sin 20°$
Therefore $\cos 20° \cos 40° \cos 80° \cdot \frac{1}{2} = \frac{1}{16}$
So, $\frac{4 \cdot \sin 20° \cos 20°}{\frac{1}{16}} = 64$
$\sqrt{3}\operatorname{cosec} 20° - \sec 20° = \frac{\sqrt{3}\cos 20° - \sin 20°}{\sin 20° \cos 20°}$
$\sqrt{3}\cos 20° - \sin 20° = 2\left(\frac{\sqrt{3}}{2}\cos 20° - \frac{1}{2}\sin 20°\right)$
$= 2\cos(20° + 30°) = 2\cos 50°$
So numerator = $\frac{2\cos 50°}{\sin 20° \cos 20°} = \frac{4\cos 50°}{\sin 40°} = 4$
Denominator:
$\cos 20° \cos 40° \cos 60° \cos 80°$ where $\cos 60° = \frac{1}{2}$
Using product formula:
$\sin 20° \cos 20° \cos 40° \cos 80° = \frac{1}{8}\sin 20°$
Therefore $\cos 20° \cos 40° \cos 80° \cdot \frac{1}{2} = \frac{1}{16}$
So, $\frac{4 \cdot \sin 20° \cos 20°}{\frac{1}{16}} = 64$
Hence, the correct answer is option: 'C'.