using the substitution x= tan a, show that
y = tan-1 2x /1+x2
reduces to
y = 2 tan-1 x
this is what i have done:-
tan y = 2x/1+x2
tan y = 2 tan a / 1 + tan2 a
tan y = 2 tan a / sec2 a
tan y = 2sin a cos a = sin 2a
i cant work this out further...can you help me plz?
first off, i think the denominator should be: 1 - x^2
secondly, you would write it like this: y = arctan [(2x)/(1 - x^2)] .......arctan means the inverse tangent. or you could write tan^{-1} [(2x)/(1 - x^2)]
Now here's how we do it:
Let
But the right hand side is the double angle formula for tangent, so we get:
Since, , we have:
as desired
You converted [1 +tan^2(a)] into sec^2(a). Although the conversion is true, it is not needed here.
2tan(a) / 1+tan^2(a)
is tan(2a).
So,
tan(y) = tan(2a)
Get the arctans both sides,
y = 2a -----------------------(i)
But x = tan(a)
So, getting the arctans of both sides,
arctan(x) = a
Substitute that into (i),
y = 2[arctan(x)] --------------shown.