anonymous
  • anonymous
Verify the identity. Justify each step. (PLEASE SHOW YOUR WORK)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{ \sec \theta }{ \csc \theta - \cot \theta } - \frac{ \sec \theta }{ \csc \theta +\cot \theta } = 2 \csc \theta \]
imqwerty
  • imqwerty
1st talke lcm and write the equation
anonymous
  • anonymous
I don't know how to do any of this

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imqwerty
  • imqwerty
What is the LCM of 1/2 and 1/3
anonymous
  • anonymous
what does lcm mean?
imqwerty
  • imqwerty
Least common multiple
anonymous
  • anonymous
2 and 3?
welshfella
  • welshfella
LCM is csc^2 O - cot^2O
anonymous
  • anonymous
ok than what
imqwerty
  • imqwerty
After taking the lcm we get the common denominator nd then we can do addition nd subtraction in the numerators.
welshfella
  • welshfella
making it one fraction we get secO(csc^O + cot^2O) - secO(csc^2O - cot^2O) ----------------------------------------- csc^2 O - cot^2O = 2 secO cot^2O ----------- csc^2 O - cot^2O
anonymous
  • anonymous
ok than what?
imqwerty
  • imqwerty
After this ^ step convert all values in tems of sin and cos
welshfella
  • welshfella
now csc^2 O = 1 + cot^2 O so the denominator = 1 so we are left with 2 sec O cot^2 O
welshfella
  • welshfella
hmm might have slipped up somewhere cant get 2 csc O from that...
welshfella
  • welshfella
yea i see where i went wrong when making it into one fraction its ecO(cscO + cot O) - secO(csc O - cot O) ---------------------------------- csc^2 O - cot^2O
welshfella
  • welshfella
= 2sec O cot O = 2 *1 /cos O * ccos O / sin O = 2 csc O
welshfella
  • welshfella
the denominator csc^2 O - cot^2O = 1 because csc^2 O = 1 + cot^2 O substituting;- 1 + cot^2 O - cot^2 O = 1
LynFran
  • LynFran
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welshfella
  • welshfella
yea - good work LynFran
anonymous
  • anonymous
So the answer is 2csctheta
anonymous
  • anonymous
@LynFran
LynFran
  • LynFran
yes when proving identities, whats on the left side of the equation must equals to the right side...
anonymous
  • anonymous
Sweet thank you!
LynFran
  • LynFran
welcome

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