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
use the relation, \(\csc^2x = 1+ \cot^2 x\)
cot^2 x is given , just plug in values to get csc^2 x!
whats 1+ 5/7 =.... ?
i got 1.71
don't use calculator, try to get answer in fraction only...
or, you can seee which option comes out to be 1.71......
i still dont understand
1+5/7 = (7+5)/7 = ... ?
it would be 6/7/=12/7
it would be 12/7, yes.
how did you get 6/7
i never got 6/7 i got, \(1+\dfrac{5}{7}= \dfrac{7+5}{7}=\dfrac{12}{7}\)
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/
1 comes from this relation, \(\huge \csc^2x = 1+ \cot^2 x\)
thats the standard relation, which you should've have memorized ...so that you can use here...
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