Give the value of tan[arctan x+1/x-1+arctan x-1/x]?

Question

I got 1/x? i don't think thats right is it
?

Answer 1

tan [arctan [(x+1)/(x-1)] + arctan([ (x - 1)/x ])

What you have to use is the tan addition identity, which goes

tan(a + b) = [tan(a) + tan(b)]/[1 - tan(a)tan(b)]

One thing to recognize is that tan and arctan are inverses (more specifically, near-inverses, since the function of tan(x) is not one-to-one) of each other, and thus cancel each other out; that is tan(arctan(a)) = a. As a result, we will have ZERO tans in our equation, and it should become

tan [arctan [(x+1)/(x-1)] + arctan([ (x - 1)/x ]) =

{ [(x+1)/(x - 1)] + [(x-1)/x] } / {1 - [(x+1)/(x-1)][(x-1)/x] }

I'll leave it up to you to solve this complex fraction.

Answer 2

Find the value of tan[arctan x+1/x-1+arctan x-1/x].

Use the trig addition formula for tangent.

tan(a + b) = [tan a + tan b]/[1 - (tan a)(tan b)]

Let

a = arctan[(x+1)/(x - 1)]
b = arctan[(x - 1)/x]

Then we have:

tan a = tan{arctan[(x+1)/(x - 1)]} = (x+1)/(x - 1)
tan b = tan{arctan[(x - 1)/x]} = (x - 1)/x

This gives us

tan(a + b) = [tan a + tan b]/[1 - (tan a)(tan b)]
tan(a + b) = [(x+1)/(x - 1) + (x - 1)/x] / {1 - [(x+1)/(x - 1)][(x - 1)/x]}
tan(a + b) = [x(x - 1) + (x - 1)^2] / [x(x - 1) - (x + 1)(x - 1)]
tan(a + b) = (2x^2 - x + 1)/(-x + 1)
tan(a + b) = (2x + 1)(x - 1)/[-(x - 1)]
tan(a + b) = -(2x + 1) = -2x - 1

Answer 3

arctanx+arctany=arctan (x+y)/(1-xy)
here arctan(x+1)/(x-1)+arctan (x-1)/x
=arctan [(x+1)/(x-1)+(x-1)/x]/1-[(x+1)/(x-1)][(x-1)/x]
simplify