Prove that cot^-1 [cos x + sinx / cos x - sin x] = pi/4 -x
lipi:
is it sinx/(cosx-sinx)
No !! :)
I have typed it right ..
by the way its an Inverse trigo function (just saying :P )
I have typed it right ..
by the way its an Inverse trigo function (just saying :P )
Oops Inverse trigo question .. sorry
and is it cot^-1*[.............]
yaa
is it [sinx+cosx]/[cosx-sinx]
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Lhs=cot^-1[(cosx+sinx)/(cosx-sinx)]
now for inverse trio function
eg there is formula
sin^-1(sinx)=x
similarly
cot^-1(cot (pi/4-x))=pi/4-x
now
cot(pi/4-x)=(cosx+sinx)/(cosx-sinx)
so
Lhs=pi/4-x
LHS=RHS
now for inverse trio function
eg there is formula
sin^-1(sinx)=x
similarly
cot^-1(cot (pi/4-x))=pi/4-x
now
cot(pi/4-x)=(cosx+sinx)/(cosx-sinx)
so
Lhs=pi/4-x
LHS=RHS
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