If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$ then $\tan x + \cot x=?$

Question

Hello :) I hit a problem. If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$, then how much is $\tan x + \cot x$?

Answer 1

We have $$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{1}{\sin x\cos x}.$$ So you want to know how much is $\sin x\cos x$.

Squaring the equality you know, we have: \begin{align*} (\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}. \end{align*}

Answer 2

HINT.

• $\tan{x}+\cot{x} = \frac{1}{\sin{x} \cdot \cos{x}}$

• $(\sin{x}+\cos{x})^{2} -1 = 2 \sin{x} \cdot \cos{x}$

Answer 3

Since $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\cdots$ (similar to tangent), you can rewrite $\tan x+\cot x$ in terms of sine and cosine. The result can be simplified to an expression involving the product of sine and cosine. The square of $\sin x+\cos x$ can also be simplified to an expression involving the product of sine and cosine. Together, this yields an answer.

However, in this particular instance, $\sin x+\cos x=\frac{\sqrt{3}+1}{2}$ strongly suggests $\sin x=\frac{\sqrt{3}}{2}$ and $\cos x=\frac{1}{2}$ (or vice versa), as that will happen for a particularly nice value of $x$, which will make calculating $\tan x+\cot x$ simple.

Answer 4

Another more general approach is to solve your equation for $x$. Since it is linear in $\sin x$ and $\cos x$ it can be transformed into a quadratic equation in $\tan \frac{x}{2}$ (see this answer):

$$\sin x+\cos x=\frac{1+\sqrt{3}}{2}\Leftrightarrow \frac{2\tan \frac{x}{2}}{% 1+\tan ^{2}\frac{x}{2}}+\frac{1-\tan ^{2}\frac{x}{2}}{1+\tan ^{2}\frac{x}{2}}% =\frac{1+\sqrt{3}}{2}$$

$$\Leftrightarrow 2\tan \frac{x}{2}+1-\tan ^{2}\frac{x}{2}=\left( 1+\tan ^{2}% \frac{x}{2}\right) \frac{1+\sqrt{3}}{2}.$$

Set $y=\tan \frac{x}{2}$

$$2y+1-y^{2}=\frac{1+\sqrt{3}}{2}+\frac{1+\sqrt{3}}{2}y^{2}\Leftrightarrow \left( 1+\frac{1+\sqrt{3}}{2}\right) y^{2}-2y-1+\frac{1+% \sqrt{3}}{2}=0$$

and solve for $y$

$$y_{1}=\frac{1}{3}\sqrt{3},y_{2}=2-\sqrt{3}$$

Hence

$$x_{1}=2\arctan \frac{1}{3}\sqrt{3}=\frac{1}{3}\pi $$

or

$$x_{2}=2\arctan \left( 2-\sqrt{3}\right) =\frac{1}{6}\pi .$$

And finally

$$\tan \frac{1}{3}\pi +\cot \frac{1}{3}\pi =\frac{4}{3}\sqrt{3}$$

or

$$\tan \frac{1}{6}\pi +\cot \frac{1}{6}\pi =\frac{4}{3}\sqrt{3}.$$

Answer 5

Hint: write $\tan(x) + \cot(x)$ in terms of $\sin(x)$ and $\cos(x)$, with a common denominator. Square $\sin(x) + \cos(x) = {\sqrt{3} + 1 \over 2}$ and compare.