# Is my textbook wrong ? (Trig identities)

The question is to evaluate $\cot 30^\circ + \cot 60^\circ$. I got $\dfrac{4}{\sqrt{3}}$. The correct answer is $\dfrac{4\sqrt{3}}{3}$. I have no idea how. Can someone please show me why I am wrong?

Both are same. $${3\over4\sqrt3}={\sqrt3\times\sqrt3\over4\sqrt3}={\sqrt3\over4}.$$

• As a side comment, the solutions manuals usually resents roots in the denominator, so they will do everything they can to avoid them (like $1/\sqrt2=\sqrt2/2$). – Arthur Aug 8 '14 at 10:08
• @Arthur I've always been taught that's the correct way to write a solution but I don't understand why. $\frac{1}{\sqrt{2}}$ or $2^{-1/2}$ seem so much neater than $\frac{\sqrt{2}}{2}$. – Jam Aug 8 '14 at 10:15
• @oliveeuler It's just a convention thing. People tend to prefer nice denominators to nice numerators. – Arthur Aug 8 '14 at 10:19
• @oliveeuler I'm not sure if this is entirely accurate, but I think that I was once told that the reason for this is because, back in the days of lookup tables and such, it was a lot easier to divide $\sqrt{2}$ by $2$ than it was to divide $1$ by $\sqrt{2}$ when you needed it to use it for something. So now it's just a convention that's lived on in mathematics curricula everywhere. – Brian Aug 8 '14 at 10:24
• @BrianScholl That makes sense. I suppose it's probably also sensible to have a standard since people often forget that $\sqrt{3}$ is a factor of $3$. – Jam Aug 8 '14 at 10:28

$\cot 30$ + $\cot 60$ = $\sqrt 3$ + $\frac{1}{\sqrt 3}$ = $\frac{(\sqrt 3*\sqrt 3) + 1}{\sqrt 3}$=$\frac{3+1}{\sqrt 3}$=$\frac{4}{\sqrt 3}$=$\frac{4*\sqrt 3}{\sqrt 3*\sqrt 3}$=$\frac{4\sqrt 3}{3}$. Got it ???

$$\cot(30^°)+\cot(60^°)=\frac{3}{\sqrt{3}}+\frac{1}{\sqrt{3}}=\frac{4}{\sqrt{3}}= \frac{4\cdot \sqrt 3}{\sqrt 3 \cdot \sqrt 3} = \frac{4\sqrt 3}{3}$$

so both are the same.