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

October 2018 2 21 Report

Cot x + tan x = cot x sec² x
(tan x+ cot x) cos x sin x-cos² x= sin² x

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}$

oktaekaputri trimakasih banyak.

Takamori37 Iya, sama-sama.

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