anonymous
  • anonymous
Help me! I will become a fan and give you a medal! (:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
with what
anonymous
  • anonymous
Hold on let me get the equation done
anonymous
  • anonymous
ok

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anonymous
  • anonymous
f your medal, and i dont need any more fans.
anonymous
  • anonymous
that made heaps of sense.
anonymous
  • anonymous
im to trilllllllllll
anonymous
  • anonymous
really ^
anonymous
  • anonymous
\[\frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc -\cot } = 2 \csc\]
ganeshie8
  • ganeshie8
\[\frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc -\cot } = 2 \csc \] ?
anonymous
  • anonymous
r u serious.... oh thats better
anonymous
  • anonymous
shut up lashaun. you a fake mf
anonymous
  • anonymous
them fighting words...
anonymous
  • anonymous
u dont even know me
anonymous
  • anonymous
give him some Lashaun
anonymous
  • anonymous
If you aren't going to help please leave
anonymous
  • anonymous
Shaylynnwalls im really sorry
ganeshie8
  • ganeshie8
\[\frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc -\cot } = 2 \csc \] left hand side is becoming 0. sure the problem doesnt have any typos ?
anonymous
  • anonymous
Its ok, it isnt your fault. Some people do not understand
anonymous
  • anonymous
And yes i am sure.
ganeshie8
  • ganeshie8
then, its a false identity !
anonymous
  • anonymous
Wait there is a typo. The second equation the bottom is csc theta plus cot theta
ganeshie8
  • ganeshie8
\[\frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc + \cot } = 2 \csc \] like this ?
anonymous
  • anonymous
yep.
anonymous
  • anonymous
I think you do have to change everything to cosine and sine
ganeshie8
  • ganeshie8
good idea, but guess that would complicate things
ganeshie8
  • ganeshie8
lets take the LCM of denominator and simplify a bit
anonymous
  • anonymous
Ok, that makes more sense
ganeshie8
  • ganeshie8
\[\frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc + \cot } = 2 \csc \] Left Hand Side : \[\frac{ \sec (\csc + \cot) - \sec ( \csc - \cot ) }{ (\csc - \cot )(\csc + \cot ) } \]
anonymous
  • anonymous
i will give u C===8 for free
ganeshie8
  • ganeshie8
simplify a bit
anonymous
  • anonymous
Ok let me write this down and than I can simplify.
ganeshie8
  • ganeshie8
\[\frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc + \cot } = 2 \csc \] Left Hand Side : \[\frac{ \sec (\csc + \cot) - \sec ( \csc - \cot ) }{ (\csc - \cot )(\csc + \cot ) } \] \[\frac{ \sec \csc + \sec \cot - \sec \csc + \sec \cot }{ (\csc - \cot )(\csc + \cot ) } \]
anonymous
  • anonymous
So it would be sec times everything on the top?
ganeshie8
  • ganeshie8
\[\frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc + \cot } = 2 \csc \] Left Hand Side : \[\frac{ \sec (\csc + \cot) - \sec ( \csc - \cot ) }{ (\csc - \cot )(\csc + \cot ) } \] \[\frac{ \sec \csc + \sec \cot) - \sec \csc + \sec \cot ) }{ (\csc - \cot )(\csc + \cot ) } \] \[\frac{ \cancel{\sec \csc} + \sec \cot) - \cancel{\sec \csc} + \sec \cot ) }{ (\csc - \cot )(\csc + \cot ) } \]
ganeshie8
  • ganeshie8
yes ! you can do that...
ganeshie8
  • ganeshie8
you do in your way on ur notes. factoring sec is a brilliant idea.. go ahead :)
ganeshie8
  • ganeshie8
\[\frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc + \cot } = 2 \csc \] Left Hand Side : \[\frac{ \sec (\csc + \cot) - \sec ( \csc - \cot ) }{ (\csc - \cot )(\csc + \cot ) } \] \[\frac{ \sec \csc + \sec \cot - \sec \csc + \sec \cot }{ (\csc - \cot )(\csc + \cot ) } \] \[\frac{ \cancel{\sec \csc} + \sec \cot - \cancel{\sec \csc} + \sec \cot }{ (\csc - \cot )(\csc + \cot ) } \] \[\frac{ 2 \sec \cot }{ (\csc^2 - \cot^2) } \]
anonymous
  • anonymous
Ok so once you do that you can simplify those to cot and csc if I did that correctly
ganeshie8
  • ganeshie8
yes ! bottom equals 1 as \(\csc^2 - \cot^2 = 1\)
ganeshie8
  • ganeshie8
\[\frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc + \cot } = 2 \csc \] Left Hand Side : \[\frac{ \sec (\csc + \cot) - \sec ( \csc - \cot ) }{ (\csc - \cot )(\csc + \cot ) } \] \[\frac{ \sec \csc + \sec \cot - \sec \csc + \sec \cot }{ (\csc - \cot )(\csc + \cot ) } \] \[\frac{ \cancel{\sec \csc} + \sec \cot - \cancel{\sec \csc} + \sec \cot }{ (\csc - \cot )(\csc + \cot ) } \] \[\frac{ 2 \sec \cot }{ (\csc^2 - \cot^2) } \] \[2 \sec \cot\] \[2 \frac{1}{\cos} \frac{\cos}{\sin} \]
anonymous
  • anonymous
Than you would simplify the bottom equation by taking out the cos?
ganeshie8
  • ganeshie8
exactly !
ganeshie8
  • ganeshie8
complete solution, go thru :- \(\large \frac{ \sec }{ \csc -\cot } -\frac{ \sec }{ \csc + \cot } = 2 \csc \) Left Hand Side : \(\large \frac{ \sec (\csc + \cot) - \sec ( \csc - \cot ) }{ (\csc - \cot )(\csc + \cot ) } \) \(\large \frac{ \sec \csc + \sec \cot - \sec \csc + \sec \cot }{ (\csc - \cot )(\csc + \cot ) } \) \(\large \frac{ \cancel{\sec \csc} + \sec \cot - \cancel{\sec \csc} + \sec \cot }{ (\csc - \cot )(\csc + \cot ) } \) \(\large \frac{ 2 \sec \cot }{ (\csc^2 - \cot^2) } \) \(\large 2 \sec \cot\) \(\large 2 \frac{1}{\cos} \frac{\cos}{\sin} \) \(\large 2\csc\)
anonymous
  • anonymous
Ok thank you so much! That makes so much sense now! The final answer would be 2csc!
ganeshie8
  • ganeshie8
np :)

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