Question 1197611
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The goal is to find the value of {{{cot(x) - tan(x)}}} based on {{{csc(x)-sec(x) = (sqrt(13))/6}}} and the restriction {{{0 < x < pi/4}}}


Note: *[tex \large 13^{1/2} = \sqrt{13}]


Let's simplify {{{cot(x) - tan(x)}}} a bit like shown below
You'll need a list of trig identities such as this here
<a href = "https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf">https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf</a>
{{{cot(x) - tan(x)}}}


{{{cos(x)/sin(x) - sin(x)/cos(x)}}}


{{{(cos^2(x))/(sin(x)cos(x)) - (sin^2(x))/(sin(x)cos(x))}}}


{{{(cos^2(x)-sin^2(x))/(sin(x)cos(x))}}}


{{{(cos(2x))/(0.5sin(2x))}}}


{{{2*cot(2x)}}}



In short, {{{cot(x) - tan(x)}}} is the same as {{{2*cot(2x)}}}


{{{cot(x) - tan(x)=2*cot(2x)}}} is an identity.
Keep this in mind for later.


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Now let's solve for x in {{{csc(x)-sec(x) = (sqrt(13))/6}}}


{{{csc(x)-sec(x) = (sqrt(13))/6}}}


{{{1/sin(x)-1/cos(x) = (sqrt(13))/6}}}


{{{cos(x)/(sin(x)cos(x))-sin(x)/(sin(x)cos(x)) = (sqrt(13))/6}}}


{{{(cos(x)-sin(x))/(sin(x)cos(x)) = (sqrt(13))/6}}}


{{{(cos(x)-sin(x))/(0.5sin(2x)) = (sqrt(13))/6}}}


{{{2(cos(x)-sin(x))/(sin(2x)) = (sqrt(13))/6}}}


{{{(2(cos(x)-sin(x))/(sin(2x)))^2 = ((sqrt(13))/6)^2}}} Squaring both sides


{{{(4(cos^2(x)-2sin(x)cos(x)+sin^2(x)))/(sin^2(2x)) = 13/36}}}


{{{(4(cos^2(x)+sin^2(x)-2sin(x)cos(x)))/(sin^2(2x)) = 13/36}}}


{{{(4(1-sin(2x)))/(sin^2(2x)) = 13/36}}}


{{{(4(1-w))/(w^2) = 13/36}}} Let w = sin(2x)


Let's solve for w.
{{{(4(1-w))/(w^2) = 13/36}}}


{{{36*4(1-w) = 13w^2}}}


{{{144-144w = 13w^2}}}


{{{13w^2+144w-144 = 0}}}
Use the quadratic formula (I'll skip showing the steps) to find that the two roots for w are
w = -12 or w = 12/13


We're told that 
0 < x < pi/4
Multiply each side by 2
2*0 < 2x < 2*pi/4
0 < 2x < pi/2
Then apply sine to each
sin(0) < sin(2x) < sin(pi/2)
0 < sin(2x) < 1


This shows that the output of sin(2x) must be between 0 and 1, which rules out w = -12 aka sin(2x) = -12
Also, the range of sin(x) is {{{-1<=sin(x)<=1}}}. So even if we didn't have to worry about 0 < x < pi/4, it's still impossible to have sin(2x) = -12 (not unless we want to involve complex numbers, but we'll stay in the real number set).


We'll ignore w = -12 and go for w = 12/13 only.


Therefore, sin(2x) = 12/13
Isolating x gets us x = 0.5*arcsin(12/13)
There are infinitely many solutions to sin(2x) = 12/13, but again we focus on the interval 0 < x < pi/4, which means we only have one solution. 


Note: arcsine is the same as inverse sine aka *[tex \large \sin^{-1}(\text{x})]


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Recall the ultimate goal is to find {{{cot(x) - tan(x)}}} which we found was equivalent to {{{2*cot(2x)}}}


Plug in x = 0.5*arcsin(12/13) and simplify
{{{2*cot(2x)}}}


{{{2*cot(2*0.5arcsin(12/13))}}}


{{{2*cot(arcsin(12/13))}}}


The question now is: how can we evaluate cot(arcsin(12/13))?


Let theta = arcsin(12/13) which rearranges to sin(theta) = 12/13
Since sine = opposite/hypotenuse, this gives us a right triangle with opposite leg 12 and hypotenuse 13.
Use the pythagorean theorem to find the adjacent leg is 5 units.
We have a 5-12-13 right triangle (this is one of the infinitely many pythagorean triples).
*[illustration Screenshot_89.png]
Tangent is the ratio of opposite/adjacent
the reciprocal is cotangent which is adjacent/opposite


So cot(theta) = adjacent/opposite = 5/12
which means,
{{{2*cot(arcsin(12/13)) = 2*(5/12) = 5/6}}}



Therefore,
{{{cot(x) - tan(x) = 5/6}}}
when {{{0 < x < pi/4}}} and {{{csc(x)-sec(x) = (sqrt(13))/6}}}



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Answer: <font color=red>5/6</font>
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