Is the Equation $ |tanx + cotx| = |tanx| + |cotx| $ True for Any Value of $x$?

[DaalChawal]

Q. Is $ |tanx + cotx| = |tanx| + |cotx| $ true for any $x?$ If it is true, then find the values of $x$.

My Working -->

Since $tanx$ and $cotx$ always have the same sign, so this holds true for any value of $x$.

 

[DaalChawal]

Okay I think this should hold true for any $x$ except $n \pi$ , $\frac{(2n+1) \pi }{2}$
Am I correct?

 

[Prove It]

In general, $\displaystyle \begin{align*} \left| a + b \right| \not\equiv \left| a \right| + \left| b \right| \end{align*}$, rather $\displaystyle \begin{align*} \left| a + b \right| \leq \left| a \right| + \left| b \right| \end{align*}$. That's called the Triangle Inequality.

 

[topsquark]

Q. Is $ |tanx + cotx| = |tanx| + |cotx| $ true for any $x?$ If it is true, then find the values of $x$.

My Working -->

Since $tanx$ and $cotx$ always have the same sign, so this holds true for any value of $x$.

You have to be a bit more careful than this, as Prove It says but you essentially have [math]\left | y + \dfrac{1}{y} \right | = |y| + \left | \dfrac{1}{y} \right |[/math], which is true for [math]y \neq 0[/math].

So, yes.

-Dan