Can you prove $1 - {cos}^{2} x = sin x cos x tan x$ $1- \cos ^ { 2} x = \sin x \cos x \tan x$ ?

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Can you prove $1 - {cos}^{2} x = sin x cos x tan x$ $1- \cos ^ { 2} x = \sin x \cos x \tan x$ ?

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Answer 1 · 2017-03-26T22:15:27

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Explanation:

Recall that $tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$ and the Pythagorean identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

From the identity it can be determined that ${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x=1-cos^2x$

$sin x cos x tan x = sin x cos x (\frac{sin x}{cos x}) = sin x sin x = {sin}^{2} x = 1 - {cos}^{2} x$ $sinxcosxtanx=sinxcosx(sinx/cosx)=sinxsinx=sin^2x=1-cos^2x$

$1 - {cos}^{2} x = 1 - {cos}^{2} x$ $1-cos^2x=1-cos^2x$

$Q.E.D.$ $"Q.E.D."$

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Can you prove $1 - {cos}^{2} x = sin x cos x tan x$ $1- \cos ^ { 2} x = \sin x \cos x \tan x$ ?

1 Answer

Explanation:

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Can you prove $1 - {cos}^{2} x = sin x cos x tan x$ $1- \cos ^ { 2} x = \sin x \cos x \tan x$ ?