Q60 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics
$\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}=$
- A$\frac{3}{4}$
- Correct answerB$1$
- C$0$
- D$\frac{1}{3}$
Solution
Since, we know,
$\tan (\mathrm{A}-\mathrm{B}) \tan (\mathrm{A}+\mathrm{B})=\frac{\tan ^2 \mathrm{~A}-\tan ^2 \mathrm{~B}}{1-\tan ^2 \mathrm{~A} \cdot \tan ^2 \mathrm{~B}}$
$\Rightarrow \tan \left(60^{\circ}-\mathrm{B}\right) \cdot \tan \left(60^{\circ}+\mathrm{B}\right)=\frac{3-\tan ^2 \mathrm{~B}}{1-3 \tan ^2 \mathrm{~B}}$
$\Rightarrow \tan \left(60^{\circ}-B\right) \cdot \tan B \tan \left(60^{\circ}+B\right) \tan 3 B\ldots(i)$
Now, $\tan 6^{\circ} \cdot \tan 42^{\circ} \cdot \tan 66^{\circ} \cdot \tan 78^{\circ}$
$=\frac{\tan 6^{\circ} \cdot \tan \left(60^{\circ}-18^{\circ}\right) \cdot \tan 18^{\circ} \cdot \tan \left(60^{\circ}+18^{\circ}\right) \cdot \tan 66^{\circ}}{\tan 18^{\circ}}$
$=\frac{\tan 6^{\circ} \cdot \tan 54^{\circ} \cdot \tan 66^{\circ}}{\tan 18^{\circ}}$ (By using (i))
$=\frac{\tan \left(60^{\circ}-6^{\circ}\right) \cdot \tan 6^{\circ} \cdot \tan \left(60^{\circ}+6^{\circ}\right)}{\tan 18^{\circ}}=\frac{\tan 18^{\circ}}{\tan 18^{\circ}}=1$
Hence, the correct answer is option: 'B'.