Question

Kuiszz

Diketahui [tex] \sf \alpha [/tex] adalah suatu variabel. Jika [tex] \rm \sf{ \sin}^{2} x= \alpha \cos x \tan x- \sin x \cos x[/tex],
Maka, hasil dari [tex] \frac{{ \alpha }^{2} \tan x \: - \: 2 {\sin }^{2} x}{ \tan x}[/tex]
adalah ...
a. 2
b. sin² x cos² x
c. tan x
d. 1
e. jawaban lain {...}

#Kuis lg besokk​

Answer 1

Verified answer

Jawaban:

d. 1

Penjelasan dengan langkah-langkah:

Trigonometri

[tex]\begin{aligned}\rm\sin^2x&=\rm\alpha\cos x\tan x-\sin x\cos \: x\\ \alpha & = \rm \: \frac{\sin^{2}x + \sin \: x \cos \: x }{ \cos x \tan \: x } \\ &\rm\cdots \: \cos x \tan \: x \: \rightarrow \: \sin \: x \\ \rm\alpha&=\rm \: \frac{\sin^{2}x + \sin \: x \cos \: x }{ \sin \: x } \\ \alpha &=\rm \frac{\sin \: x(\sin \: x + \cos \: x)}{\sin \: x} \\ \alpha &=\rm\sin \: x + \cos \: x\\ { \alpha }^{2} & = \rm \: (\sin \: x + \cos \: x {)}^{2} \\ { \alpha }^{2} &= \rm(\sin \: x + \cos \: x)(\sin \: x + \cos \: x)\\ { \alpha }^{2} & = \rm \: 1 + \sin(2x) \end{aligned}[/tex]

[tex]\therefore[/tex] Maka

hasil dari [tex] \frac{{ \alpha }^{2} \tan x \: - \: 2 {\sin }^{2} x}{ \tan x}[/tex]

[tex]\displaystyle\rm=\frac{ (1+\sin (2x))\tan x \: - \: 2 {\sin }^{2} x}{ \tan x}[/tex]

[tex]\displaystyle\rm=\frac{ \tan x + \sin(2x) \tan(x) \: - \: 2 {\sin }^{2} x}{ \tan x}[/tex]

[tex]\displaystyle\rm = \frac{ \tan(x) + 2 \sin(x) \cos(x) \frac{ \sin(x) }{ \cos(x) } - 2 \sin^{2} (x) }{ \tan(x) } [/tex]

[tex] \displaystyle\rm= \frac{ \tan(x) + 2 \sin ^{2} (x )- 2 \sin^2(x) }{ \tan(x) } [/tex]

[tex] \displaystyle\rm= \frac{ \tan(x) }{ \tan(x) } [/tex]

[tex] \displaystyle\rm= 1[/tex]