Trigonometry I

We Have $\displaystyle \tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{1}{\sin x\cos x}$

So We Want to Know How Much Is $\displaystyle \sin x cosx$

Squaring the Equality You Know, We Have:

$\begin{aligned} (\sin x + \cos x)^2 &= \left(\frac{\sqrt{3}+1}{2}\right)^2\\ \sin^2 x + 2\sin x\cos x + \cos^2 x & = \frac{4 + 2\sqrt{3}}{4}\\ 1 + 2\sin x \cos x &= 1 + \frac{\sqrt{3}}{2}\\ 2\sin x \cos x &= \frac{\sqrt{3}}{2}\\ \sin x \cos x &= \frac{\sqrt{3}}{4}\\ \tan x + \cot x = \frac{1}{\sin x\cos x} &= \frac{4\sqrt{3}}{3} \rightarrow \frac{\sqrt{48}}{3} . \end{aligned}$

\displaystyle a=48 &there4 b=0 &there4 c=3

$\displaystyle a+b+c = 48+0+3 = \boxed{51}$

$\sin x+\cos x=\frac{\sqrt3+1}2=\frac{\sqrt3}2+\frac12$

But I know a value of $x$ such that $\sin x=\frac{\sqrt3}2$ and $\cos x=\frac12$ . Hm, I forget that value of $x$ , oh well. Hm, I also forget if it was $\tan x$ or $\cot x$ that was $\frac{\sin x}{\cos x}$ , but I remember that $\cot x=\frac1{\tan x}$ . No worries, I know that one of $\tan x$ or $\cot x$ is $\sqrt3$ and the other is $\frac{\sqrt3}3$ . But wait, that doesn't matter since we're adding them! $\tan x+\cot x=\frac{4\sqrt3}3$ . Whew, excuse my ignorance.

The equation $\displaystyle\ tan(x) + cot(x)$ is symmetrical in sine and cosine so ,we can assign them individual values, without loss of generality. Let $\displaystyle\ sinx = \frac{\sqrt{3}}{2} \ , \ cosx = \frac{1}{2}$ . This leaves $tanx + cotx = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{\sqrt{48}}{3}\therefore\ a+b+c = \boxed{51}$

sin x + cos x = (root 3 + 1)/2 = (root 3)/2 +1/2, so if sin x = (root 3)/2 and cos x = 1/2, so x can be taken as being equal to 60 degrees. Then you can substitute the value of x in tan x + cot x and than solve it.