Pleast help me with the problem.

Simplify the following trigonometric expression

Multiply cosθ/cscθ+1 by cscθ-1/cscθ-1

Thank you! :)

KarmaKizuna
Jul 6, 2017

#1**+3 **

**Simplify the following trigonometric expression**

**Multiply cosθ/cscθ+1 by cscθ-1/cscθ-1**

\(\begin{array}{|rcll|} \hline && \left( \frac{ \cos(\phi) } {\csc(\phi)+1 } \right) \cdot \left( \frac{ \csc(\phi)-1 } { \csc(\phi)-1 } \right ) \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\csc(\phi)-1 \Big) } { \csc^2(\phi)-1 } \quad & | \quad \csc(\phi) = \frac{1}{\sin(\phi)} \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big) } { \frac{1}{\sin^2(\phi)}-1 } \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big) } { \frac{1-\sin^2(\phi)}{\sin^2(\phi)} } \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { 1-\sin^2(\phi) } \quad & | \quad 1-\sin^2(\phi) = \cos^2(\phi) \\\\ &=& \frac{ \cos(\phi)\cdot \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { \cos^2(\phi) } \\\\ &=& \frac{ \Big(\frac{1}{\sin(\phi)}-1 \Big)\cdot \sin^2(\phi) } { \cos(\phi) } \\\\ &=& \frac{ \Big(\frac{1-\sin(\phi)}{\sin(\phi)}\Big)\cdot \sin^2(\phi) } { \cos(\phi) } \\\\ &=& \frac{ \Big(1-\sin(\phi)\Big)\cdot \sin(\phi) } { \cos(\phi) } \\\\ &=& \Big(1-\sin(\phi)\Big) \cdot \tan(\phi) \\\\ &=& \tan(\phi)-\tan(\phi)\cdot \sin(\phi) \\ \hline \end{array} \)

heureka
Jul 6, 2017

#2**+1 **

Hello Heureka,

Pleast check your message center. Thank you for helping me solve the problem. :)

KarmaKizuna
Jul 6, 2017

#3**+2 **

\(\dfrac{\cos \theta}{\csc\theta +1}\cdot \dfrac{\csc\theta - 1}{\csc\theta - 1}\\ =\dfrac{\cos \theta}{\csc\theta + 1}\\ =\dfrac{\sin 2\theta}{2(1+\sin \theta)}\)

The second fraction just have the same numerator and denominator...

MaxWong
Jul 6, 2017

#4**+1 **

Hello Max,

I got the solution to the problem. It is tanθ - tanθ sinθ

I did not know the method used to solve this problem.

KarmaKizuna
Jul 6, 2017

#5**0 **

Hi KarmaKizuna,

**Welcome to Web2.0calc forum, I hope you learn lots here :)**

You can write people private messages if you want to but please use it as a way to direct them to your query on the forum.

**Make sure you include the link to your question or the post you want them to answer or discuss.**

The question itself should be public, I mean it should be on the forum.

It seems you queried Heureka's answer privately. I would have like to have seen what that query was.

When you ask a question, it is there for us all to answer, or learn from, so please do not hide it away in a private message :)

Melody
Jul 6, 2017

#6**+2 **

Going from Max's answer, we have

sin 2θ / [2 (1 + sinθ)]

2sinθcosθ / [2 ( 1 + sin θ) ]

sinθcosθ / (1 + sinθ)

sinθcosθ ( 1 - sinθ) / [1 - sin^2θ]

sinθcosθ [ 1 - sinθ] / cos^2θ

sinθ[1 - sinθ]/cosθ

sinθ/cosθ - sin^2θ/cosθ

tanθ - tanθsinθ

The moral here is that with these trig expressions, there may not be just one "correct" answer !!!!

CPhill
Jul 6, 2017