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Answer is :-\[-\sqrt{1-x^2}\]
cos inv 0
is pi/2
ya I also did that
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use circular inverse formula ... so it is called (i haven't tried this formula though)
I haven't heard of circular inverse formula!
\[ \cos^{-1}A + \cos^{-1}B = AB + AB/(A^2+B^2) \] not sure though ... i made it up myself!!
but I have a formula in my book:- \[\cos^{-1} (A)+\cos^{-1} (B)=\cos^{-1} [AB-(\sqrt{1-A^2})(\sqrt{1-B^2})]\]
Oh ... that's the formula .. sorry!!
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np:)
but u r like a teacher
no ... not a teacher ... just stupid old guy!!
no! for me u r like a teacher:)
so what should I do now about that formula?
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should I use that?
lol ... thanks for compliment :) ... though not deserved!! put the value of A=x and B=0
k!
wait a minute plz:)
I got |dw:1338122446950:dw|
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