Yang mau dpat amal Jariyah,, Bantuin lagi..
Buktikan identitas trigonometri !!
(1 + tan²x ) (1 - cos²x) = tan²x
Buktikan identitas trigonometri !!
(1 + tan²x ) (1 - cos²x) = tan²x
Takamori37
Dengan mengingat identitas:
![$\begin{align}(1+\tan^2x)(1-\cos^2 x)&=\tan^2 x \\ \left[1+\frac{\sin^2 x}{\cos^2 x}\right]\times \sin^2 x&=\tan^2 x \\ \frac{\cos^2x+\sin^2x}{\cos^2x}\times\sin^2 x&=\tan^2 x \\ \frac{(\sin^2x+\cos^2x)}{\cos^2x}\times\sin^2x&=\tan^2x \\ \frac{1}{\cos^2x}\times\sin^2x&=\tan^2x \\ \frac{\sin^2 x}{\cos^2x}&=\tan^2x \\ \left[\frac{\sin x}{\cos x}\right]^2&=\tan^2x \\ \tan^2x=\tan^2x \\ Q.E.D \end{align}](https://tex.z-dn.net/?f=%24%5Cbegin%7Balign%7D%281%2B%5Ctan%5E2x%29%281-%5Ccos%5E2+x%29%26%3D%5Ctan%5E2+x+%5C%5C+%5Cleft%5B1%2B%5Cfrac%7B%5Csin%5E2+x%7D%7B%5Ccos%5E2+x%7D%5Cright%5D%5Ctimes+%5Csin%5E2+x%26%3D%5Ctan%5E2+x+%5C%5C+%5Cfrac%7B%5Ccos%5E2x%2B%5Csin%5E2x%7D%7B%5Ccos%5E2x%7D%5Ctimes%5Csin%5E2+x%26%3D%5Ctan%5E2+x+%5C%5C+%5Cfrac%7B%28%5Csin%5E2x%2B%5Ccos%5E2x%29%7D%7B%5Ccos%5E2x%7D%5Ctimes%5Csin%5E2x%26%3D%5Ctan%5E2x+%5C%5C+%5Cfrac%7B1%7D%7B%5Ccos%5E2x%7D%5Ctimes%5Csin%5E2x%26%3D%5Ctan%5E2x+%5C%5C+%5Cfrac%7B%5Csin%5E2+x%7D%7B%5Ccos%5E2x%7D%26%3D%5Ctan%5E2x+%5C%5C+%5Cleft%5B%5Cfrac%7B%5Csin+x%7D%7B%5Ccos+x%7D%5Cright%5D%5E2%26%3D%5Ctan%5E2x+%5C%5C+%5Ctan%5E2x%3D%5Ctan%5E2x+%5C%5C+Q.E.D+%5Cend%7Balign%7D "$\begin{align}(1+\tan^2x)(1-\cos^2 x)&=\tan^2 x \\ \left[1+\frac{\sin^2 x}{\cos^2 x}\right]\times \sin^2 x&=\tan^2 x \\ \frac{\cos^2x+\sin^2x}{\cos^2x}\times\sin^2 x&=\tan^2 x \\ \frac{(\sin^2x+\cos^2x)}{\cos^2x}\times\sin^2x&=\tan^2x \\ \frac{1}{\cos^2x}\times\sin^2x&=\tan^2x \\ \frac{\sin^2 x}{\cos^2x}&=\tan^2x \\ \left[\frac{\sin x}{\cos x}\right]^2&=\tan^2x \\ \tan^2x=\tan^2x \\ Q.E.D \end{align}")
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Thanks 2
Yuniia
ku kira cara ku salah.. ternyata bener, gak ada yg slah.. mksih lagi yaa :D
Takamori37
Oke
cherlyaja23
qed t ap y