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

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The equation |tanx + cotx| = |tanx| + |cotx| is true for any value of x except at points where tanx or cotx are undefined, specifically at nπ and (2n+1)π/2. Since tanx and cotx share the same sign for all valid x, the equation holds true in those ranges. The discussion references the Triangle Inequality, which states that |a + b| is less than or equal to |a| + |b|, but in this case, equality is achieved when both terms are non-zero. Therefore, the equation is valid as long as x is not at the points where tanx or cotx are undefined. This confirms the equation's validity under specified conditions.
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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 $x$.
 
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Okay I think this should hold true for any $x$ except $n \pi$ , $\frac{(2n+1) \pi }{2}$
Am I correct?
 
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.
 
DaalChawal said:
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
 
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