How do you prove $sec (x) - cos (x) = sin (x) \cdot tan (x)$ $Sec(x) - cos(x) = sin(x) * tan(x)$ ?

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How do you prove $sec (x) - cos (x) = sin (x) \cdot tan (x)$ $Sec(x) - cos(x) = sin(x) * tan(x)$ ?

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Answer 1 · 2016-07-10T07:59:11

Knowing that $sec (x) = \frac{1}{cos (x)}$ $sec(x) = 1/cos(x)$ and $tan (x) = \frac{sin (x)}{cos (x)}$ $tan(x) = sin(x)/cos(x)$ , rewriting the equation yields

$\frac{1}{cos (x)} - \frac{cos (x)}{1} = sin (x) \cdot (\frac{sin (x)}{cos (x)})$ $1/cos(x) - cos(x)/1 = sin(x) * (sin(x)/cos(x))$

Rewriting the left fraction by using the property

$\frac{a}{b} - \frac{c}{d} = \frac{a d - b c}{b d}$ $a/b-c/d =(ad-bc)/(bd)$

and simplifying the right side of the equation yields

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

Note that in this case, we can make use of the identity

${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x)=1$ , since $1 - {cos}^{2} (x) = {sin}^{2} (x)$ $1-cos^2(x)=sin^2(x)$ , giving us

$\frac{{sin}^{2} (x)}{cos (x)} = \frac{{sin}^{2} (x)}{cos (x)}$ $sin^2(x)/cos(x)=sin^2(x)/cos(x)$ , which is true, therefore we have proven that

$sec (x) - cos (x) = sin (x) tan (x)$ $sec(x)-cos(x) = sin(x) tan(x)$

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How do you prove $sec (x) - cos (x) = sin (x) \cdot tan (x)$ $Sec(x) - cos(x) = sin(x) * tan(x)$ ?

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How do you prove Sec(x) - cos(x) = sin(x) * tan(x)sec(x)−cos(x)=sin(x)⋅tan(x)Sec(x) - cos(x) = sin(x) * tan(x)?

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How do you prove $sec (x) - cos (x) = sin (x) \cdot tan (x)$ $Sec(x) - cos(x) = sin(x) * tan(x)$ ?