Question

The value of $(\sqrt{3}+\tan 1^{\circ })(\sqrt{3}+\tan 2^{\circ })(\sqrt{3}+\tan 3^{\circ })\cdots \cdots (\sqrt{3}+\tan {28}^{\circ })(\sqrt{3}+\tan {29}^{\circ })$ is

• A. $2^{28}$
• B. $2^{15}+2-\sqrt{3}$
• C. $2^{29}$
• D. $2^{14}+2-\sqrt{3}$

Answer 1

The correct option is C $2^{29}$

Given : $(\sqrt{3}+\tan 1^{\circ })(\sqrt{3}+\tan 2^{\circ })(\sqrt{3}+\tan 3^{\circ })\cdots \cdots (\sqrt{3}+\tan {28}^{\circ })(\sqrt{3}+\tan {29}^{\circ })$

We know that
$\sqrt{3}+\tan ({30}^{\circ }-x)=\sqrt{3}+\frac{\tan {30}^{\circ }-\tan x}{1+\tan {30}^{\circ }\cdot \tan x}=\sqrt{3}+\frac{1-\sqrt{3}\tan x}{\sqrt{3}+\tan x}=\frac{4}{\sqrt{3}+\tan x}$

Now,
$(\sqrt{3}+\tan 1^{\circ })(\sqrt{3}+\tan {29}^{\circ })=\frac{4}{\sqrt{3}+\tan {29}^{\circ }}\times (\sqrt{3}+\tan {29}^{\circ })=4$
Similarly,
$(\sqrt{3}+\tan 2^{\circ })(\sqrt{3}+\tan {28}^{\circ })=4$

So,
$(\sqrt{3}+\tan 1^{\circ })(\sqrt{3}+\tan 2^{\circ })(\sqrt{3}+\tan 3^{\circ })\cdots (\sqrt{3}+\tan {28}^{\circ })(\sqrt{3}+\tan {29}^{\circ })=4^{14}\cdot (\sqrt{3}+\tan {15}^{\circ })=4^{14}\cdot (\sqrt{3}+\frac{\sqrt{3}-1}{\sqrt{3}+1})=4^{14}\cdot \left(\frac{2+2\sqrt{3}}{\sqrt{3}+1}\right)=2^{29}$