If cot^2x=5/7, then csc^2 is equivalent to
a)12/7
b)7/5
c)5sqrt(74)/74
d)sqrt(74)/7
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hartnn (hartnn):
use the relation,
\(\csc^2x = 1+ \cot^2 x\)
hartnn (hartnn):
cot^2 x is given , just plug in values to get csc^2 x!
hartnn (hartnn):
whats 1+ 5/7 =.... ?
OpenStudy (anonymous):
i got 1.71
hartnn (hartnn):
don't use calculator,
try to get answer in fraction only...
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hartnn (hartnn):
or, you can seee which option comes out to be 1.71......
OpenStudy (anonymous):
i still dont understand
hartnn (hartnn):
1+5/7 = (7+5)/7 = ... ?
OpenStudy (anonymous):
it would be 6/7/=12/7
hartnn (hartnn):
it would be 12/7, yes.
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OpenStudy (anonymous):
how did you get 6/7
hartnn (hartnn):
i never got 6/7
i got, \(1+\dfrac{5}{7}= \dfrac{7+5}{7}=\dfrac{12}{7}\)
OpenStudy (anonymous):
and how did you get the 1 there.sorry if im askin so much questions im trying to understand.when i read the question sometimes i dont know how to start them or dnt know what they want to do when it comes to math/
hartnn (hartnn):
1 comes from this relation,
\(\huge \csc^2x = 1+ \cot^2 x\)
hartnn (hartnn):
thats the standard relation, which you should've have memorized ...so that you can use here...