- #1

FritoTaco

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## Homework Statement

__: [itex]csc(tan^{-1}\dfrac{x}{2})=\sqrt{\dfrac{x^{2}+4}{x}}[/itex]__

Problem 1

Problem 1

__Problem 2__: [itex]\sqrt{\dfrac{1-sinx}{1+sinx}}=\dfrac{|cosx|}{1+sinx}[/itex]

## Homework Equations

**Quotient Identities**[itex]tan\theta=\dfrac{sin\theta}{cos\theta}[/itex]

[itex]cos\theta=\dfrac{cos\theta}{sin\theta}[/itex]

**Pythagorean Identites**[itex]sin^{2}\theta+cos^{2}=1[/itex]

[itex]cot^{2}\theta+1=csc^{2}\theta[/itex]

[itex]tan^{2}+1=sec^{2}\theta[/itex]

**Recirpocal Identites**[itex]sin\theta=\dfrac{1}{csc\theta}[/itex]

[itex]cos\theta=\dfrac{1}{sec\theta}[/itex]

[itex]tan\theta=\dfrac{1}{cot\theta}[/itex]

[itex]csc\theta=\dfrac{1}{sin\theta}[/itex]

[itex]sec\theta=\dfrac{1}{cos\theta}[/itex]

[itex]cot\theta=\dfrac{1}{tan\theta}[/itex]

**(maybe useful?)**

Cofunction IdentitiesCofunction Identities

[itex]sin(\dfrac{\pi}{2}-\theta)=cos\theta[/itex]

[itex]cos(\dfrac{\pi}{2}-\theta)=sin\theta[/itex]

[itex]tan(\dfrac{\pi}{2}-\theta)=cot\theta[/itex]

[itex]csc(\dfrac{\pi}{2}-\theta)=sec\theta[/itex]

[itex]sec(\dfrac{\pi}{2}-\theta)=csc\theta[/itex]

[itex]cot(\dfrac{\pi}{2}-\theta)=tan\theta[/itex]

*: [itex]\sqrt{x^{2}}=|x|[/itex]*

**Note**## The Attempt at a Solution

**Problem 1:**Any ideas on how to get started? I have no idea what to do with inverse tangent and that stupid x/2 and it makes me mad! I know I need some work for it to be a valid thread so here goes nothing.

[itex]csc(tan^{-1}\dfrac{x}{2})=\sqrt{\dfrac{x^{2}+4}{x}}[/itex]

Maybe manipulate the left side?

[itex]csc(cot\theta)[/itex]

...yup, I'm totally lost.

**Problem 2:**

[itex]\sqrt{\dfrac{1-sinx}{1+sinx}}=\dfrac{|cosx|}{1+sinx}[/itex]

I manipulated the left side

[itex]\dfrac{\sqrt{1-sinx}}{\sqrt{1+sinx}}\cdot\dfrac{\sqrt{1-sinx}}{\sqrt{1-sinx}}[/itex]

[itex]\dfrac{1-sinx}{\sqrt{(1+sinx)(1-sinx)}}[/itex]

[itex]\dfrac{1-sinx}{\sqrt{(1+sin^{2}x)}} <--\sqrt{x^{2}}=|x|[/itex]

[itex]\dfrac{1-sinx}{|cosx|}[/itex]

[itex]\dfrac{|cosx|}{1+sinx}[/itex] Am I allowed to just switch it around like that?

[itex]\dfrac{|cosx|}{1+sinx}[/itex] = [itex]\dfrac{|cosx|}{1+sinx}[/itex]