If sin + tan=5/6 Find sin*tan

[man111]

if [tex]\sin x+\tan x = \frac{5}{6},[/tex] then [tex]\sin x\tan x[/tex] is

[nevrodermit]

Нека [tex]\sin{(x)}\tan{(x)}=a\Rightarrow \frac{\sin^2{(x)}}{\cos{(x)}}=a\Rightarrow \sin^2{(x)}=a\cos{(x)}\Rightarrow 1-\cos^2{(x)}=a\cos{(x)}\Rightarrow \cos^2{(x)}=1-a\cos{(x)}(1)[/tex].
Повдигаме на втора степен [tex]\sin{(x)}+\tan{(x)}=\frac{5}{6}\Leftrightarrow \sin{(x)}+\frac{\sin{(x)}}{\cos{(x)}}=\frac{5}{6}\Leftrightarrow \sin^2{(x)}+\frac{\sin^2{(x)}}{\cos^2{(x)}}+2a=\frac{25}{36}[/tex][tex]\Leftrightarrow 1-\cos^2{(x)}+\frac{1-\cos^2{(x)}}{\cos^2{(x)}}+2a=\frac{25}{36}\Leftrightarrow 1-\cos^2{(x)}+\frac{1}{\cos^2{(x)}}-1+2a=\frac{25}{36}[/tex][tex]\Leftrightarrow -cos^2{(x)}+\frac{1}{\cos^2{(x)}}+2a=\frac{25}{36}\Leftrightarrow -1+a\cos{(x)}+\frac{1}{1-a\cos{(x)}}=\frac{25}{36}\Leftrightarrow \frac{-(1-a\cos{(x)})^2+1}{1-a\cos{(x)}}=\frac{25}{36}[/tex][tex]\Leftrightarrow \frac{2a\cos{(x)}-a^2\cos^2{(x)}}{1-a\cos{(x)}}=\frac{25}{36}\Leftrightarrow \frac{2a\cos{(x)}-a^2\cos^2{(x)}}{\cos^2{(x)}}=\frac{25}{36}\Leftrightarrow \frac{2a-a^2\cos{(x)}}{\cos{(x)}}=\frac{25}{36}[/tex][tex]\Leftrightarrow 72a-36a^2\cos{(x)}=25\cos{(x)}\Leftrightarrow \cos{(x)}(25+36a^2)=72a\Leftrightarrow \cos{x}=\frac{72}{25+36a^2}[/tex].
Заместваме в (1) и получаваме [tex]\cos^2{(x)}=1-\frac{72a^2}{25+36a^2}=\frac{25-36a^2}{25+36a^2}[/tex]. Оттук идва уравнението
[tex]\frac{(72a)^2}{(25+36a^2)^2}=\frac{25-36a^2}{25+36a^2}\Leftrightarrow (72a)^2=25^2-36^2a^4[/tex]. Полагаме [tex]t=a^2[/tex] и получаваме [tex]72^2t=25^2-36^2t^2\Leftrightarrow 36^2t^2+72^2t-25^2=0[/tex]. Дискриминанта [tex]D=72^4+4.36^2.25^2=2^6.3^4.37.157[/tex]. Корени [tex]t_{1,2}=\frac{-72^2\pm 72\sqrt{37.157}}{2.36^2}=\frac{-72\pm \sqrt{5809}}{36}[/tex]. От формулите на Виет единият корен е отрицателен, затова [tex]a^2=\frac{-72+ \sqrt{5809}}{36}\Rightarrow a=\pm \sqrt{\frac{-72+ \sqrt{5809}}{36}}=\pm \frac{\sqrt{-72+ \sqrt{5809}}}{6}[/tex].