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The problem is as follows:

First assume that $x$ is on:

$\frac{3\pi}{2}<x<2\pi$

Then find the least integer value of the expression from below:

$$\cot x + \frac{\csc^2x+\cot ^4x}{\csc^2x+\cot x}+\tan x + \frac{\sec^2x+\tan^4x}{\sec^2x+\tan x}$$

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{5}\\ 2.&\textrm{2}\\ 3.&\textrm{4}\\ 4.&\textrm{3}\\ \end{array}$

I am not sure how to tackle this problem. What I've attempted to do is to attempt to build the square.

This is seen in below:

$$\cot x +\frac{1+\cot^2 x + \cot ^4 x}{1+\cot^2x+\cot x}+\tan x + \frac{1+\tan^2 x + \tan^4x }{1+\tan^2x+\tan x}$$

This reduces to:

$$\cot x +\frac{\left(\frac{1}{2}+\cot^2 x\right)^2 +\frac{3}{4}}{\left(\cot x+\frac{1}{2}\right)^2+\frac{3}{4}}+\tan x + \frac{\left(\frac{1}{2}+\tan^2 x\right)^2+\frac{3}{4}}{\left(\frac{1}{2}+\tan x\right)^2+\frac{3}{4}}$$

But that's it this is the part where I got stuck. I don't know what else can be done here in order to simplify the expression?. You can notice that the expressions in the denominator are of a lesser power.

Does it exist any sort of algebraic manipulation which I'm overlooking? Can someone help me here please?

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$$\sec^2 x=1+\tan^2 x \, , \quad \csc^2 x=1+\cot^2 x$$

Use these identities to rewrite the rational terms as

$$\frac{\sec^2x+\tan^4x}{\sec^2x+\tan x}=\frac{\tan^4x+\tan^2 x +1}{\tan^2x+\tan x+1}$$

Now use the identity : $$t^4+t^2+1=(t^2+t+1)(t^2-t+1)$$

The original expression simplifies $$\cot x + (\cot^2 x -\cot x +1) + \tan x + (\tan^2 x - \tan x +1)$$ $$=\cot^2 x + \tan^2 x +2$$ $$=(\cot^2 x + \tan^2 x -2)+4$$ $$=(\cot x - \tan x)^2+4$$

This will be minimum when first term is zero. There exists a value of $x \in (3\pi/2, 2\pi)$ for which $\cot x =\tan x$ (namely $7\pi/4$). Thus the minimum attained is $4$.

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    $\begingroup$ In case the OP or anyone is curiuos about the identity $$t^4+t^2+1=(t^2-t+1)(t^2+t+1)$$ its because $$t^4+t^2+1=t^4+2t^2+1-t^2={(t^2+1)}^2-t^2$$ $\endgroup$ – Albus Dumbledore Feb 5 at 9:03
  • $\begingroup$ @cosmo5 I'm stuck with the last part. How can I get the minimum?. My best guess was to use $AM\geq GM$, hence $\frac{\cot x + \tan x}{2}\geq \sqrt{\tan x \cot x}$ thus $\cot x + \tan x \geq 2$ but since it is squared then $\left(\cot x + \tan x\right)^2 \geq 4$ thus the minimum would be $4$ alternative $3$. But does it exist another method?. Can you please include this as a special request in your answer?, so I can accept it. $\endgroup$ – Chris Steinbeck Bell Feb 5 at 9:20
  • $\begingroup$ @ChrisSteinbeckBell See the edit. $\endgroup$ – cosmo5 Feb 5 at 9:27
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    $\begingroup$ @cosmo5 Gee I never thought I had to use an algebraic manipulation to get a difference and then squaring this. Probably this is the suggested approach when facing these situations. I believe those functions have been chosen particulary for this problem. Interesting. Thanks buddy. $\endgroup$ – Chris Steinbeck Bell Feb 5 at 9:41
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The key is to use the cubic identities $1-u^3 = (1-u)(1+u+u^2), 1+u^3 = (1+u)(1-u+u^2)$. For convenience, let $\cot x = c$, $\tan x = t$; substitute $c^2$ and $t^2$ into the first identity to get:

$$c + \frac{1+c^2+c^4}{1+c+c^2} + t + \frac{1+t^2+t^4}{1+t+t^2}$$ $$= c+t + \frac{1 - c^6}{1 - c^2} \frac{1 - c}{1 - c^3} + \frac{1 - t^6}{1 - t^2} \frac{1-t}{1 - t^3}$$ $$= c+t + \frac{(1-c^3)(1+c^3)}{(1-c)(1+c)} \frac{1 - c}{1 - c^3} + \frac{(1 - t^3)(1+t^3)}{(1-t)(1+t)} \frac{1-t}{1 - t^3}$$ $$= c+t + 1-c+c^2 + 1-t+t^2 = \tan^2 x + \cot^2 x + 2$$ $$= (\tan x + 1/\tan x)^2 - 2 + 2 ≥ 4$$

where in the last step, square both sides of $u+1/u < -2$ when $x < 0, x \in \mathbb R$ ($\tan x$ is negative in the given range).

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