Question

Solve:
$tan^{-1} (2+x) + tan^{-1} (2-x)$ = $tan ^{-1} 2/3$

Answer 1

I'm going to use arctan as tan^(-1). arctan(2-x) + arctan (2+x) =arctan [(2-x) + (2 +x)]/( 1- (2-x)(2+x)) =arctan(4)/(x^2-3) arctan(4)/(x^2 - 3) = arctan(2/3) 4/(x^2-3) = 2/3 12/2 = (x^2 - 3) 6= x^2 - 3 9 = x^2 x=+9 or -9

Answer 2

As we know that,
tan⁻¹a +tan⁻¹b = tan⁻¹[(a+b)/(1-ab)]

So , tan⁻¹(x+2)+tan⁻¹(2-x)
=tan⁻¹{[(2+x) +(2-x)]/[1-(2+x)(2-x)]}
=tan⁻¹{4/[1-(4-x²)]}
=tan⁻¹[4/(x²-3)]

and since,
tan⁻¹(x+2)+tan⁻¹(2-x)= tan⁻¹(2/3)
so,tan⁻¹[4/(x²-3)]= tan⁻¹(2/3)

equating both sides,
4/(x²-3) = 2/3
⇒x²-3 =6
⇒x² = 9
⇒x = 3 or -3