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?

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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?

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it should be \[ \frac{1}{\sin x + 1} + \frac{1}{\csc x +1} = 1 \] \[ \frac{\csc x + 1 + \sin x + 1}{( \sin x + 1)( \csc x + 1)} = 1\]
well first write csc in terms of sin :-) that will be easy :-)
\[ \frac{1}{\sin x + 1} + \frac{1}{\rm \color{reD}{\frac{1}{sinx} +1}} = 1 \] like this solve red part first

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So \[ \frac{\frac{1}{sin x} x + 1 + \sin x + 1}{( \sin x + 1)( \frac{1}{sin x} x + 1)}\]
better to gget rid of fraction first :-)
so do I times the numerator and denominator by sin??
\[ \frac{1}{\sin x + 1} + \frac{1}{\rm \color{reD}{\frac{1}{sinx} +1}} = 1 \] \[\large\rm \color{reD}{\frac{ 1 }{ \sin x } +1}\] multiply by sinx/1 do you know how to find common denominator ? just for the red part
For the red part the common denominator is sin x and 1. For the red we have \[ 1+ \sin x \] corect?
for the red part |dw:1434223526447:dw| common denominator is sin x
yes correct. I forgot to include the denominator
in other words multiply by sinx/1 |dw:1434223578712:dw| let me know if you have any question :-)
Got it
alright move on
\[ \huge\rm \frac{1}{\sin x + 1} + \frac{1}{\rm \color{reD}{\frac{1 + sinx }{sinx} }} = 1 \] now change divison to multiplication \[\huge\rm \frac{ 1 }{ \frac{ 1+sinx }{ sinx } }\] multiply top with the reciprocal of the bottom fraction
for example \[\large\rm \frac{ a }{ \frac{ b }{ c } } = a \times \frac{ c }{ b }\]
so is it \( \frac{1}{1} \)
so \( \frac{sin }{1+ sin x} \)
yes right :-)
\[ \huge\rm \frac{1}{\sin x + 1} + \frac{sin x}{\rm \color{reD}{ sinx +1 }} = 1 \] 1 + sinx is same as sinx +1
so we have \[ \frac{sin x}{sinx + 1} = 1 \] but that does not equal 1]
check numerator again that's not right
I see... Right in front of my face... lol man these things make me see crosseye
so we have \[ \frac{1+sin x}{sinx + 1}\ = 1 but that does not equal 1]
why not ??
so we have \[ \frac{1+sin x}{sinx + 1} = 1 \]
1+ sin x is same as sinx+1
That post before my last post was from me copying and pasting and I did not take the last part out.
Thank you. I was working it all types of ways but all I had to do was work the bottom denominator . Thank you!!
my pleasure :-) gO_od job!! :-)

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