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anonymous
 one year ago
Trig/ Pre Cal/identies Am I on the right path?
Please do not give answer but guide me. Thank you.
The problem
\[ \frac{1}{\sin x + 1} + \frac{1}{\csc x +1} \]
\[ \frac{\csc x + 1 + \sin x + 1}{( \sin x + 1)( \csc x + 1)}\]
Am I on the right path?
anonymous
 one year ago
Trig/ Pre Cal/identies Am I on the right path? Please do not give answer but guide me. Thank you. The problem \[ \frac{1}{\sin x + 1} + \frac{1}{\csc x +1} \] \[ \frac{\csc x + 1 + \sin x + 1}{( \sin x + 1)( \csc x + 1)}\] Am I on the right path?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it should be \[ \frac{1}{\sin x + 1} + \frac{1}{\csc x +1} = 1 \] \[ \frac{\csc x + 1 + \sin x + 1}{( \sin x + 1)( \csc x + 1)} = 1\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1well first write csc in terms of sin :) that will be easy :)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{1}{\sin x + 1} + \frac{1}{\rm \color{reD}{\frac{1}{sinx} +1}} = 1 \] like this solve red part first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So \[ \frac{\frac{1}{sin x} x + 1 + \sin x + 1}{( \sin x + 1)( \frac{1}{sin x} x + 1)}\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1better to gget rid of fraction first :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do I times the numerator and denominator by sin??

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{1}{\sin x + 1} + \frac{1}{\rm \color{reD}{\frac{1}{sinx} +1}} = 1 \] \[\large\rm \color{reD}{\frac{ 1 }{ \sin x } +1}\] multiply by sinx/1 do you know how to find common denominator ? just for the red part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the red part the common denominator is sin x and 1. For the red we have \[ 1+ \sin x \] corect?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1for the red part dw:1434223526447:dw common denominator is sin x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes correct. I forgot to include the denominator

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1in other words multiply by sinx/1 dw:1434223578712:dw let me know if you have any question :)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[ \huge\rm \frac{1}{\sin x + 1} + \frac{1}{\rm \color{reD}{\frac{1 + sinx }{sinx} }} = 1 \] now change divison to multiplication \[\huge\rm \frac{ 1 }{ \frac{ 1+sinx }{ sinx } }\] multiply top with the reciprocal of the bottom fraction

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1for example \[\large\rm \frac{ a }{ \frac{ b }{ c } } = a \times \frac{ c }{ b }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is it \( \frac{1}{1} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \( \frac{sin }{1+ sin x} \)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[ \huge\rm \frac{1}{\sin x + 1} + \frac{sin x}{\rm \color{reD}{ sinx +1 }} = 1 \] 1 + sinx is same as sinx +1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we have \[ \frac{sin x}{sinx + 1} = 1 \] but that does not equal 1]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1check numerator again that's not right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see... Right in front of my face... lol man these things make me see crosseye

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we have \[ \frac{1+sin x}{sinx + 1}\ = 1 but that does not equal 1]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we have \[ \frac{1+sin x}{sinx + 1} = 1 \]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.11+ sin x is same as sinx+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That post before my last post was from me copying and pasting and I did not take the last part out.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you. I was working it all types of ways but all I had to do was work the bottom denominator . Thank you!!

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1my pleasure :) gO_od job!! :)
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