Question Number 139521 by mnjuly1970 last updated on 28/Apr/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:#\:\:\:{nice}\:…\:{calculus}# \\ $$$$\:\:\:\:\:\:\:\:{if}\:\:\:\:\left({pq}\right)^{{sin}^{\mathrm{2}} \left({x}\right)} +\left({pq}\right)^{{cos}^{\mathrm{2}} \left({x}\right)} ={p}+{q} \\ $$$$\:\:\:\:\:\:\:\:\:\:{then}\:\:\:\:{tan}\left({x}\right)=? \\ $$
Answered by qaz last updated on 28/Apr/21
$$\ll\mathrm{1}\gg:{p}={q}=\mathrm{0}\:\:\:\mathrm{tan}\:{x}=??? \\ $$$$\ll\mathrm{2}\gg:{p}={q}=\mathrm{1}\:\:\:\mathrm{tan}\:{x}=??? \\ $$$$\ll\mathrm{3}\gg:{p},{q}\in\left(\mathrm{0},\mathrm{1}\right)\cup\left(\mathrm{1},\infty\right),{pq}\neq\mathrm{1} \\ $$$$\left({pq}\right)^{\mathrm{sin}\:^{\mathrm{2}} {x}} +\left({pq}\right)^{\mathrm{cos}\:^{\mathrm{2}} {x}} \\ $$$$=\left({pq}\right)^{\mathrm{sin}\:^{\mathrm{2}} {x}} +\left({pq}\right)^{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}} ={p}+{q} \\ $$$$\Rightarrow\:\:\:\mathrm{sin}\:^{\mathrm{2}} {x}={log}_{\left({pq}\right)} {p}\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} {x}={log}_{\left({pq}\right)} {q} \\ $$$${or}\:\:\:\mathrm{sin}\:^{\mathrm{2}} {x}={log}_{\left({pq}\right)} {q}\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} {x}={log}_{\left({pq}\right)} {p} \\ $$$$\mathrm{tan}\:{x}=\sqrt{{log}_{{p}} {q}}\:\:\:\:\:{or}\:\:\sqrt{{log}_{{q}} {p}} \\ $$$$\ll\mathrm{4}\gg:{p},{q}\in{C}\:\:\:\mathrm{tan}\:{x}=??? \\ $$$$….. \\ $$