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OpenStudy (anonymous):
sec (tan^-1(-4) + sin^-1(1/3))
I know tangent is in the IV quadrant. How can I tell what sin and cos are?
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OpenStudy (ankit042):
why is tangent in IV quadrant?
OpenStudy (ankit042):
sin is +ve in 1st and second quadrant
OpenStudy (anonymous):
because it can only occur in the first our fourth qudrant
OpenStudy (anonymous):
the doman is -pi/2<x<pi/2
OpenStudy (anonymous):
thats the domain for the inverse function of tangent
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OpenStudy (ankit042):
Hmm that is true! same way you can do for inverse of sin it's range is also in between -pi/2 to pi/2
OpenStudy (anonymous):
so is tangent 8/2 or is it 16/4 i mean i guess it doesnt matter because the proportions would all be the same
OpenStudy (ankit042):
Just use the calculator to get the angles and proceed not sure what is the doubt
OpenStudy (anonymous):
can you simplify sqrt68
OpenStudy (anonymous):
2sqrt 17
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OpenStudy (anonymous):
got it
OpenStudy (anonymous):
I believer using cos(A+B) = cos(A)cos(B) - sin(A)sin(B) will do the trick
OpenStudy (anonymous):
why is that
OpenStudy (anonymous):
and then put 1/cos for the sec value?
OpenStudy (anonymous):
yes
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