Find an angle $A$ such that $\tan A + \cot A = 2$

Question

Find an angle $A$ such that $\tan A + \cot A = 2$

I'm recently getting into trigonometry, so i think I may have some trouble, that's why I'm asking.

I used the identities $$\tan A = \frac{\sin A}{\cos A}$$ and $$\cot A = \frac{\cos A}{\sin A}$$

So it follows $$\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin^2 A+\cos^2 A}{(\sin A)(\cos A)}$$ and we have that $\sin^2 A+\cos^2 A =1$, so it clearly follows that $$(\sin A)(\cos A)=\frac{1}{2}$$

From here, is there any identity i could use?.

What i did next was to try to use the basis of sine and cosine, so if we let $a$ be the opposite and $b$ the adyacent, we have $$(\sin A)(\cos A)=\frac{ab}{h^2}=\frac{ab}{a^2+b^2}=\frac{1}{2}$$ And thus it's possible to find the answer if we find positive values for $a,b$, but i don't have more ideas.

Answer 1

Note that

• $\cot A = \frac{1}{\tan A}$
• $x + \frac{1}{x} \geq 2$ for $x>0$ and
• $x + \frac{1}{x} \color{blue}{=} 2 \Leftrightarrow \color{blue}{x= 1}$

It follows, a possible solution is $$\tan A = 1 \Rightarrow A = \frac{\pi}{4}$$

Answer 2

$\sin(a)\cos(a) = \dfrac{\sin(2a)}{2}$

So clearly one solution is $a = \dfrac{\pi}{4}$. All solutions are of the form $a=\pi n+ \dfrac{\pi}{4}$ for integer $n$.

Answer 3

Since people have already answered, just for an intuition and idea what is going on, I recommend you to look at the graph of the function.

The blue color line represents the RHS of the equation and red color graph represents $\cot(x)+\tan(x)$.

The point where these two curve intersect is the angle you want.

As other have said, $x=\pi/4$ works and as you can see in the graph there are infinitely many choices of angle you have here.

Have fun exploring and playing with graphical visualization!! https://www.desmos.com/calculator/8gnfia2ifz