Prove-that-tan-1-p-p-2q-tan-1-q-p-q-pi-4-

Question Number 9941 by Tawakalitu ayo mi last updated on 17/Jan/17

Prove that .

\tan^{- 1} [\frac{p}{p + 2 q}] + \tan^{- 1} [\frac{q}{p + q}] = \frac{π}{4}

Answered by mrW1 last updated on 17/Jan/17

l e t α = \tan^{- 1} [\frac{p}{p + 2 q}] a n d β = \tan^{- 1} [\frac{q}{p + q}]

l e t u = \tan^{- 1} [\frac{p}{p + 2 q}] + \tan^{- 1} [\frac{q}{p + q}] = α + β

\tan u = \tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β}

= \frac{\frac{p}{p + 2 q} + \frac{q}{p + q}}{1 - \frac{p}{p + 2 q} \times \frac{q}{p + q}} = \frac{p (p + q) + q (p + 2 q)}{(p + 2 q) (p + q) - p q}

= \frac{p^{2} + p q + p q + 2 q^{2}}{p^{2} + 2 p q + p q + 2 q^{2} - p q} = \frac{p^{2} + 2 p q + 2 q^{2}}{p^{2} + 2 p q + 2 q^{2}} = 1

\Rightarrow u = \tan^{- 1} 1 = \frac{π}{4} + n π, n \in Z

Commented by Tawakalitu ayo mi last updated on 17/Jan/17

Thank you sir . God bless you .