# Math Help - Range of trigonometric expression?

1. ## Range of trigonometric expression?

[(3 - tan^2 x)/(tan^2 x)]*[(1 - tan^2 x)/(tan^2 x)]*[(2 tan x)/(1 - tan^2 x)]*[(1 - 3 tan^2 x)/(3 tan x - tan^3 x)] - [3cot^2 x - 1]*[cot^2 x - 1] is the given expression.

Prove this expression is less than or equal to 1.

I couldn't prove it to be less than or equal to 1.(Maybe I messed up in the intermediate steps where after simplification I got the expression = -[(cot^2 x + 1)]^2

2. That's quite a mess.

Your final expression is less than or equal to zero (0). Squaring makes positive.

That is either sufficient to the task or you wandered off somwehere.

3. How is the final expression less than or equal to one? Please explain.

4. Originally Posted by fardeen_gen
[(3 - tan^2 x)/(tan^2 x)]*[(1 - tan^2 x)/(tan^2 x)]*[(2 tan x)/(1 - tan^2 x)]*[(1 - 3 tan^2 x)/(3 tan x - tan^3 x)] - [3cot^2 x - 1]*[cot^2 x - 1] is the given expression.

Prove this expression is less than or equal to 1.

I couldn't prove it to be less than or equal to 1.(Maybe I messed up in the intermediate steps where after simplification I got the expression = -[(cot^2 x + 1)]^2
it cannot be that since that is equal to -csc^4 x, right?

i'll put my simplification until some point..

$
\frac{3-\tan^2 x}{\tan^2 x} \cdot \frac{1-\tan^2 x}{\tan^2 x} \cdot \frac{2\tan x}{1-\tan^2 x} \cdot \underbrace{\frac{1-3\tan^2 x}{3\tan -\tan^3 x}}_{\tan x(3-\tan^2 x)} - \frac{3\cot^2 x-1}{\cot^2 x-1} \cdot \underbrace{\frac{\tan^2x}{\tan^2x}}_{=1}$
(we assume that $x\not=0$)

$= \frac{2(1-3\tan^2 x)}{(\tan^4 x)(1-\tan^2 x)}-\frac{3-\tan^2x}{1-\tan^2 x} = \frac{2(1-3\tan^2 x)-(\tan^4x)(3-\tan^2x)}{(\tan^4 x)(1-\tan^2 x)}$

5. Is the problem wrong then?

6. Hello,

Originally Posted by kalagota
it cannot be that since that is equal to -csc^4 x, right?

i'll put my simplification until some point..

$
\frac{3-\tan^2 x}{\tan^2 x} \cdot \frac{1-\tan^2 x}{\tan^2 x} \cdot \frac{2\tan x}{1-\tan^2 x} \cdot \underbrace{\frac{1-3\tan^2 x}{3\tan -\tan^3 x}}_{\tan x(3-\tan^2 x)} - \frac{3\cot^2 x-1}{\cot^2 x-1} \cdot \underbrace{\frac{\tan^2x}{\tan^2x}}_{=1}$
(we assume that $x\not=0$)

$= \frac{2(1-3\tan^2 x)}{(\tan^4 x){\color{red}(1-\tan^2 x)}}-\frac{3-\tan^2x}{1-\tan^2 x} = \dots$
Isn't there a typo here ? (in red) it shouldn't be here.

This gives $\frac{\tan^6x+3\tan^4x-8\tan^2x+2}{\tan^4x(1-\tan^2x)}$

7. $- \frac{3\cot^2 x-1}{\cot^2 x-1} \cdot \underbrace{\frac{\tan^2x}{\tan^2x}}_{=1}" alt="- \frac{3\cot^2 x-1}{\cot^2 x-1} \cdot \underbrace{\frac{\tan^2x}{\tan^2x}}_{=1}" />

The expression was - [3 cot^2 x -1]. [cot^2 x - 1]

8. Originally Posted by fardeen_gen
The expression was - [3 cot^2 x -1]. [cot^2 x - 1]
He just said he multiplied by tan²/tan², which is possible because it's equal to 1.

For the simplification, I get to $-1+\frac{4}{1-\tan^2x}+\frac{8}{\tan^4x}-\frac{6}{\tan^4x(1-\tan^2x)}$

9. $- \frac{3\cot^2 x-1}{\cot^2 x-1}$
I could understand multiplying by tan^2 x/tan^2 x. But the part of the expression was not a fraction. It was a product.( ie (3 cot^2 x - 1) multiplied by (cot^2 x - 1) ,not divided by (cot^2 x - 1))

God save us! It seems the problem is jinxed. The expression simplify refuses to behave and simply into a reasonable quantity.

10. Originally Posted by fardeen_gen
I could understand multiplying by tan^2 x/tan^2 x. But the part of the expression was not a fraction. It was a product.( ie (3 cot^2 x - 1) multiplied by (cot^2 x - 1) ,not divided by (cot^2 x - 1))

God save us! It seems the problem is jinxed. The expression simplify refuses to behave and simply into a reasonable quantity.
Oh sorry !

Let's take it from scratch.

$\frac{3-\tan^2 x}{\tan^2 x} \cdot \frac{1-\tan^2 x}{\tan^2 x} \cdot \frac{2\tan x}{1-\tan^2 x} \cdot \underbrace{\frac{1-3\tan^2 x}{3\tan -\tan^3 x}}_{\tan x(3-\tan^2 x)} - (3\cot^2 x-1)(\cot^2 x-1)$

Simplifying the left term and multiplying the right term by $\frac{\tan^4x}{\tan^4x}$, we get :

$\frac{2(1-3\tan^2x)}{\tan^4x}-\frac{[\tan^2x (3 \cot^2x-1)][\tan^2x(\cot^2x-1)]}{\tan^4x}$

$=\frac{2(1-3 \tan^2x)-(3-\tan^2x)(1-\tan^2x)}{\tan^4x}$

$=\frac{2-6\tan^2x-3+4 \tan^2x-\tan^4x}{\tan^4x}$

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

$=\boxed{-\frac{(\tan^2x+1)^2}{\tan^4x}}$

And this is always less than 1 because it's negative. This is only true if $x \neq 0 \bmod \pi$
So you will have to find the value if x=0.