Prove that $sec x - cos x = tan x sin x$ $secx-cosx=tanxsinx$ ?

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Answer 1 · 2016-10-08T15:51:57

Please see below.

$sec x - cos x$ $secx-cosx$

= $\frac{1}{cos x} - cos x$ $1/cosx-cosx$

= $\frac{1 - {cos}^{2} x}{cos x}$ $(1-cos^2x)/cosx$

= $\frac{{sin}^{2} x}{cos x}$ $sin^2x/cosx$

= $\frac{sin x}{cos x} \times sin x$ $sinx/cosx xxsinx$

= $tan x sin x$ $tanxsinx$

Answer 2 · 2016-10-08T15:55:19

I believe you have made an error in the question as:

$sec (x) - cos (x) \equiv \frac{1}{cos x} - cos x$ $sec(x)-cos(x)-=1/cosx-cosx$

$:.sec(x)-cos(x) -=1/cosx-cosx*cosx/cosx$

$:.sec(x)-cos(x) -=1/cosx-cos^2x/cosx$

$:.sec(x)-cos(x) -=(1-cos^2x)/cosx$

$:.sec(x)-cos(x) -=(sin^2x)/cosx$ (using $sin^2x+cos^2x-=1$ )

$:.sec(x)-cos(x) -=sinx/cosx*sinx$

$:.sec(x)-cos(x) -=tanx*sinx$

Prove that secx-cosx=tanxsinxsecx−cosx=tanxsinxsecx-cosx=tanxsinx?