Question

Prove the identity. [Hint: Let u = tan−1(x) and v = tan−1(y), so that x = tan(u) and y = tan(v). Use an Addition Formula to find tan(u + v).] (Enter your answers in terms of u and v.) tan−1 x + y 1 − xy = tan−1(x) + tan−1(y), −1 < x < 1, −1 < y < 1

Answer 1

Answer with Step-by-step explanation:

LHS

$tan^{-1}x+tan^{-1}y$

$u=tan^{-1}x, v=tan^{-1}y$

$tan(u+v)=\frac{tanu+tanv}{1-tanutanv}$

By using the formula

$tan(x+y)=\frac{tanx+tany}{1-tanxtany}$

Substitute the values

$tan(tan^{-1}x+tan^{-1}y)=\frac{tan(tan^{-1}x)+tan(tan^{-1}y)}{1-tan(tan^{-1}x)tan(tan^{-1}y)}$

$tan(tan^{-1}x+tan^{-1}y)=\frac{x+y}{1-xy}$

$tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy})$

Hence, proved.