Trolling Trignometry

I'm assuming that you used the method of AM-GM as per:

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$\begin{aligned} 3\sin x + 3\cos x + 3\tan x + 3\cot x + 3 \sec x + 3\csc x + 6 & = 3 \left( \sin x + \csc x \right) + 3 \left( \cos x + \sec x \right) + 3 \left( \tan x + \cot x \right) + 6 \\ & = 3 \left( \sin x + \dfrac{1}{\sin x} \right) + 3 \left( \cos x + \dfrac{1}{\cos x} \right) + 3 \left( \tan x + \dfrac{1}{\tan x} \right) + 6 \\ & \ge 3(2) + 3(2) + 3(2) + 6 ~~~~~~~~~~~~~ \boxed{\small\color{#3D99F6}{\text{As}~ a+\frac{1}{a} \ge 2}} \\ & \ge 24 \end{aligned}$

Hence, minimum value of expression is $\boxed{24}$ .

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But this approach is wrong.

$\bigstar$ There are certain things that you forgot to consider (even I forgot when I solved it):

• The value of $x$ that gives $\sin x + \dfrac{1}{\sin x} = 2$ is $n \pi + {(-1)}^{n} \dfrac{\pi}{2}$ . Whereas, the value of $x$ that gives $\cos x + \dfrac{1}{\cos x} = 2$ is $2n \pi$ and the value of $x$ that gives $\tan x + \dfrac{1}{\tan x} = 2$ is $n \pi + \dfrac{\pi}{4}$ . Hence, you can see that the set of all common solutions of $x$ satisfying the above three conditions is $\phi$ .

• The most important mistake that most forget to consider is that AM-GM can only be applied to positive real numbers. Considering the fact that you've asked for the minimum value of the expression, we can go to negative reals as well. If we went to negative reals as well, then we find that the minimum value for the expression is $-\infty$ .

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Set of questions that can replace the above problem:

Q1. Find the minimum value of $\left| 3\sin x + 3\cos x + 3\tan x + 3\cot x + 3 \sec x + 3\csc x + 6 \right|$ .

Ans. $3+6\sqrt2$ . For the solution, click here .

Q2. Find the minimum value of $3\sin x + 3\cos y + 3\tan z + 3\cot z + 3 \sec y + 3\csc x + 6$ , given that $x,y$ and $z$ lie in the first quadrant.

Ans. $24$ . Using AM-GM.

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I have published a report as well.