Question Number 9941 by Tawakalitu ayo mi last updated on 17/Jan/17
![Prove that. tan^(−1) [(p/(p + 2q))] + tan^(−1) [(q/(p + q))] = (π/4)](Q9941.png)
$$\mathrm{Prove}\:\mathrm{that}. \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{p}}{\mathrm{p}\:+\:\mathrm{2q}}\right]\:+\:\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{q}}{\mathrm{p}\:+\:\mathrm{q}}\right]\:=\:\frac{\pi}{\mathrm{4}} \\ $$
Answered by mrW1 last updated on 17/Jan/17
![let α=tan^(−1) [(p/(p + 2q))] and β=tan^(−1) [(q/(p + q))] let u=tan^(−1) [(p/(p + 2q))] + tan^(−1) [(q/(p + q))] =α+β tan u=tan (α+β)=((tan α+tan β)/(1−tan α tan β)) =(((p/(p+2q))+(q/(p+q)))/(1−(p/(p+2q))×(q/(p+q))))=((p(p+q)+q(p+2q))/((p+2q)(p+q)−pq)) =((p^2 +pq+pq+2q^2 )/(p^2 +2pq+pq+2q^2 −pq))=((p^2 +2pq+2q^2 )/(p^2 +2pq+2q^2 ))=1 ⇒u=tan^(−1) 1=(π/4)+nπ, n∈Z](Q9943.png)
$${let}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{p}}{\mathrm{p}\:+\:\mathrm{2q}}\right]\:{and}\:\beta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{q}}{\mathrm{p}\:+\:\mathrm{q}}\right]\: \\ $$$${let}\:{u}=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{p}}{\mathrm{p}\:+\:\mathrm{2q}}\right]\:+\:\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{q}}{\mathrm{p}\:+\:\mathrm{q}}\right]\:=\alpha+\beta \\ $$$$\mathrm{tan}\:{u}=\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta} \\ $$$$=\frac{\frac{{p}}{{p}+\mathrm{2}{q}}+\frac{{q}}{{p}+{q}}}{\mathrm{1}−\frac{{p}}{{p}+\mathrm{2}{q}}×\frac{{q}}{{p}+{q}}}=\frac{{p}\left({p}+{q}\right)+{q}\left({p}+\mathrm{2}{q}\right)}{\left({p}+\mathrm{2}{q}\right)\left({p}+{q}\right)−{pq}} \\ $$$$=\frac{{p}^{\mathrm{2}} +{pq}+{pq}+\mathrm{2}{q}^{\mathrm{2}} }{{p}^{\mathrm{2}} +\mathrm{2}{pq}+{pq}+\mathrm{2}{q}^{\mathrm{2}} −{pq}}=\frac{{p}^{\mathrm{2}} +\mathrm{2}{pq}+\mathrm{2}{q}^{\mathrm{2}} }{{p}^{\mathrm{2}} +\mathrm{2}{pq}+\mathrm{2}{q}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{u}=\mathrm{tan}^{−\mathrm{1}} \mathrm{1}=\frac{\pi}{\mathrm{4}}+{n}\pi,\:{n}\in\mathbb{Z} \\ $$
Commented by Tawakalitu ayo mi last updated on 17/Jan/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$