We know for all right triangles that:
sin x = ( o p p h y p ) {\displaystyle \sin x=\left({\frac {opp}{hyp}}\right)}
tan x = ( o p p a d j ) {\displaystyle \tan x=\left({\frac {opp}{adj}}\right)}
From the latter we can deduce
tan x × a d j = o p p {\displaystyle \tan x\times adj=opp}
Through substitution into our first equation, it follows that
sin x = ( tan x × a d j h y p ) {\displaystyle \sin x=\left({\frac {\tan x\times adj}{hyp}}\right)}
And through standard algebra:
sin x × h y p = tan x × a d j {\displaystyle \sin x\times hyp=\tan x\times adj}
tan x = ( sin x × h y p a d j ) {\displaystyle \tan x=\left({\frac {\sin x\times hyp}{adj}}\right)}
Substitute that back into sin x = ( tan x × a d j h y p ) {\displaystyle \sin x=\left({\frac {\tan x\times adj}{hyp}}\right)}
