Question:
If $\tan \theta=\sqrt{3}$, then $\sec \theta=?$
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) 2
Solution:
(d) 2
Let us first draw a right $\triangle \mathrm{ABC}$ right angled at $\mathrm{B}$ and $\angle \mathrm{A}=\theta$.
Given: $\tan \theta=\sqrt{3}$
But $\tan \theta=\frac{B C}{A B}$
So, $\frac{B C}{A B}=\frac{\sqrt{3}}{1}$
Thus, $B C=\sqrt{3} k$ and $A B=k$
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
$\Rightarrow \mathrm{AC}^{2}=(\sqrt{3} k)^{2}+(k)^{2}$
$\Rightarrow \mathrm{AC}^{2}=4 k^{2}$
$\Rightarrow A C=2 k$
$\therefore \sec \theta=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{2 k}{k}=\frac{2}{1}$