# Thread: csc-1x in terms of tan-1

1. ## csc-1x in terms of tan-1

Problem is csc-1x in terms of tan-1. And i cant really get a grip of it.
Tried it in two ways which are both wrong.
First attempt csc-1(x)=tan-1(y)=z
Thus csc(z)=x=1/sin(z) and sin(z)=1/x
tan(z)=y=sin(z)/cos(z)=sin(z)/(1-sin^2(z))^1/2) At this point i think it all gets messed up. substituting in the values for sinz=1/x gives me
y=1/((x^2-1))^1/2.
Thus tan^-1(1/((x^2-1))^1/2)=csc-1(x)=z. Which is incorrect.
And when things get wrong you try other things so here is my other attempt
csc-1(x)=sin-1(1/x)=pi/2 - cos-1(1/x))=tan-1(y)=z. Thus
csc(z)=1/sin(z)=1/(pi/2 - cos(z))=x
And tan(z)=y=sin(z)/cos(z). Substituting the values gives me y=2/(pi*x - 2). Performing the inverse and I end up with tan-1(2/(pi*x - 2))=z=csc-1(x).
Again the solution is wrong.

Correct solution is pi/2sgnx tan-1((x^2-1)^1/2)

And i dont know how to get there. Thankful if anyone can point out where im thinking wrong so I can correct it for future problems.

2. If $\displaystyle 0\le \theta\le \pi/2$ then imagine a right triangle having one angle $\displaystyle \theta$, the leg opposite that side x and the hypotenuse 1. Then $\displaystyle csc(\theta)= x/1= x$ or $\displaystyle csc^{-1}(x)= \theta$. By the Pythagorean theorem, the side opposite angle $\displaystyle \theta$ has length $\displaystyle (1- x^2)^{1/2}$. Now $\displaystyle tan(\theta)= tan(csc^{-1}(x))= \frac{(1- x^2)^{1/2}}{x}$ so that $\displaystyle csc^{-1}(x)= tan^{-1}\left(\frac{(1- x^2)^{1/2}}{x}\right)$. The "$\displaystyle \frac{\pi}{2}- sgn(x )( )$" is for angles in other quadrants.

3. Hello, Zamzen!

I don't understand their answer . . .

$\displaystyle \text{Problem: \:write }\csc^{\text{-}1}x\text{ in terms of }\tan^{\text{-}1}x.$

$\displaystyle \text{Correct solution: }\:\dfrac{\pi}{2} - \text{sgn}\,x\,\tan^{\text{-}1}\!\left(\dfrac{1}{\sqrt{x^2-1}}\right)$ .??

$\displaystyle \text{Let: }\:\theta \:=\:\csc^{\text{-}1}x \quad\Rightarrow\quad \csc\theta \:=\:\dfrac{x}{1} \:=\:\dfrac{hyp}{opp}$

$\displaystyle \,\theta$ is in a right triangle with: $\displaystyle opp = 1,\;hyp = x$
. . Pythagorus says: .$\displaystyle adj = \sqrt{x^2-1}$

$\displaystyle \text{Then: }\:\tan\theta \;=\;\dfrac{opp}{adj} \;=\;\dfrac{1}{\sqrt{x^2-1}}$

. . $\displaystyle \text{Hence: }\;\theta \;=\;\tan^{-1}\!\left(\dfrac{1}{\sqrt{x^2-1}}\right)$

$\displaystyle \text{Therefore: }\;\csc^{\text{-}1}x \;=\;\tan^{\text{-}1}\!\left(\dfrac{1}{\sqrt{x^2-1}}\right)$

4. I think you are both wrong, but I am thankful for youre efforts. I plugged it in on graphing calculator and the results you gave me do not overlap with csc-1(x).
And when i compared the solution in the book, its graph did not overlap either. I managed to find the solution which I think is correct since it overlaps with csc-1(x).
csc-1(x)=z
Thus csc(z)=x
1+1/tan^2(z)=csc^2(z)
Thus (1+1/tan^2(z))^1/2=csc(z)=x
squaring both sides and cross multiplying and i get 1/(x^2-1)=tan^2(z)
square root of both sides and +-1/(x^2-1)^1/2=tan(z). Thus tan-1(sgn(x)/(x^2-1)^1/2)=z=csc-1(x)=
sgn(x)tan-1(1/(x^2-1)^1/2).
This one overlaps with the csc-1(x) graph.