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Question Number 139521 by mnjuly1970 last updated on 28/Apr/21

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$if{\left(pq\right)}^{{sin}^2\left(x\right)}+{\left(pq\right)}^{{cos}^2\left(x\right)}=p+q$$$$thentan\left(x\right)=?$$

Answered by qaz last updated on 28/Apr/21

$$\ll 1\gg :p=q=0\tan x=???$$$$\ll 2\gg :p=q=1\tan x=???$$$$\ll 3\gg :p,q\in (0,1)\cup (1,\mathrm{\infty }),pq\ne 1$$$${\left(pq\right)}^{\sin {}^{2}x}+{\left(pq\right)}^{\cos {}^{2}x}$$$$={\left(pq\right)}^{\sin {}^{2}x}+{\left(pq\right)}^{1-\sin {}^{2}x}=p+q$$$$\Rightarrow \sin {}^{2}x={log}_{\left(pq\right)}p\cos {}^{2}x={log}_{\left(pq\right)}q$$$$or\sin {}^{2}x={log}_{\left(pq\right)}q\cos {}^{2}x={log}_{\left(pq\right)}p$$$$\tan x=\sqrt{{log}_{p}q}or\sqrt{{log}_{q}p}$$$$\ll 4\gg :p,q\in C\tan x=???$$$$.....$$

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