Cot x + tan x = cot x sec² x (tan x+ cot x) cos x sin x-cos² x= sin² x.... Question from @oktaekaputri

Takamori37 Asumsi pembuktian.
Atas.
$$\begin{align}\cot x+\tan x&=\cot x\sec^2x \\ \frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}&=\cot x\sec^2x \\ \frac{\cos^2x+\sin^2x}{\sin x\cos x}&=\cot x\sec^2x \\ \frac{1}{\sin x\cos x}&=\cot x\sec^2x \\ \frac{\cos x}{\sin x\cos^2x}&=\cot x\sec^2x \\ \frac{\cos x}{\sin x}\times\frac{1}{\cos^2x}&=\cot x\sec^2x \\ \cot x\sec^2x&=\cot x\sec^2x\end{align}$

Bawah
$$\begin{align}(\tan x+\cot x)\cos x\sin x-\cos^2x&=\sin^2x \\ \text{Pada sebelumnya, berlaku juga:} \\ \boxed{\frac{1}{\sin x\cos x}}\times\cos x\sin x-\cos^2x&=\sin^2x \\ 1-\cos^2x&=\sin^2x \\ (\sin^2x+\cos^2x)-\cos^2x&=\sin^2x\\\sin^2x&=\sin^2x\end{align}$

1 votes Thanks 2

oktaekaputri trimakasih banyak.

Takamori37 Iya, sama-sama.

izharulhaq78 $cotX + tanX = cotX sec^{2}X \\ \frac{cosX}{sinX} + \frac{sinX}{cosX} = \frac{cosX}{sinX}( \frac{1}{cos^{2}X}) \\ \frac{cos^{2}X + sin^{2}X}{cosXsinX}=\frac{1}{sinX}( \frac{1}{cosX}) \\ \frac{1}{cosXsinX}=\frac{1}{sinXcosX}$

$(tanX+cotX)cosXsinX-cos^{2}X = sin^{2}X \\ ( \frac{sinX}{cosX}+ \frac{cosX}{sinX} )cosXsinX-cos^{2}X = sin^{2}X \\ ( \frac{sin^{2}X+cos^{2}X}{cosXsinX})cosXsinX-cos^{2}X = sin^{2}X \\ (sin^{2}X+cos^{2}X)-cos^{2}X = sin^{2}X \\ sin^{2}X= sin^{2}X$

Maaf kalau salah

2 votes Thanks 2

oktaekaputri terimakasih banyak

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