Show-that-tan-1-p-p-2q-tan-1-p-p-q-pi-2-

Question Number 24554 by tawa tawa last updated on 20/Nov/17

Show that: tan^(−1) ((p/(p + 2q))) + tan^(−1) ((p/(p + q))) = (π/2)

$$\mathrm{Show}\:\mathrm{that}:\:\:\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{p}}{\mathrm{p}\:+\:\mathrm{2q}}\right)\:+\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{p}}{\mathrm{p}\:+\:\mathrm{q}}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$

Commented by mrW1 last updated on 21/Nov/17

that′s not true! let′s say p=q=1 LHS=tan^(−1) (1/3)+tan^(−1) (1/2)≈0.785≠(π/2) for p=0, LHS=0+0≠(π/2) please check your question!

$${that}'{s}\:{not}\:{true}! \\ $$$${let}'{s}\:{say}\:{p}={q}=\mathrm{1} \\ $$$${LHS}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{785}\neq\frac{\pi}{\mathrm{2}} \\ $$$${for}\:{p}=\mathrm{0}, \\ $$$${LHS}=\mathrm{0}+\mathrm{0}\neq\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$${please}\:{check}\:{your}\:{question}! \\ $$

Commented by tawa tawa last updated on 21/Nov/17

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{what}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\frac{\pi}{\mathrm{4}} \\ $$

Commented by mrW1 last updated on 21/Nov/17

I have shown that the result of LHS is no constant, not π/2, not π/4, not everything!

$${I}\:{have}\:{shown}\:{that}\:{the}\:{result}\:{of}\:{LHS} \\ $$$${is}\:{no}\:{constant},\:{not}\:\pi/\mathrm{2},\:{not}\:\pi/\mathrm{4},\:{not} \\ $$$${everything}! \\ $$

Commented by tawa tawa last updated on 21/Nov/17