Q152 / 270 · NDA 2013 (Phase 2) · Mathematics
What is $\tan 15^{\circ}$ equal to?
- Correct answerA$2-\sqrt{3}$
- B$2+\sqrt{3}$
- C$1-\sqrt{3}$
- D$1+\sqrt{3}$
Solution
From formula
$\begin{aligned}
\tan (\mathrm{A}-\mathrm{B}) & =\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A} \cdot \tan \mathrm{~B}} \\
\tan \left(15^{\circ}\right) & =\tan \left(45^{\circ}-30^{\circ}\right) \\
& =\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}+\tan 30^{\circ}} \\
& =\frac{1-\frac{1}{\sqrt{3}}}{1+1 \cdot \frac{1}{\sqrt{3}}}=\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)} \times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}
\end{aligned}$
$\begin{aligned} & =\frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2-(1)^2}=\frac{3+1-2 \sqrt{3}}{3-1} \\ & =\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3}\end{aligned}$
Hence, the correct answer is option: 'A'.