Prove that


If $\tan \theta=\sqrt{3}$, then $\sec \theta=?$

(a) $\frac{2}{\sqrt{3}}$

(b) $\frac{\sqrt{3}}{2}$

(c) $\frac{1}{2}$

(d) 2



(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

$\Rightarrow \mathrm{AC}^{2}=(\sqrt{3} k)^{2}+(k)^{2}$

$\Rightarrow \mathrm{AC}^{2}=4 k^{2}$

$\Rightarrow A C=2 k$

$\therefore \sec \theta=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{2 k}{k}=\frac{2}{1}$



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