Given that cosx=p Where x us an acute angle in degrees find in term of p i) Sin x ii)tanx iii)tan(90-x)?

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Given that cosx=p Where x us an acute angle in degrees find in term of p i) Sin x ii)tanx iii)tan(90-x)?

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Answer 1 · 2018-04-29T06:27:01

$sin x = \sqrt{1 - p^{2}}$ $sinx=sqrt(1-p^2)$

$tan x = \frac{\sqrt{1 - p^{2}}}{p}$ $tanx=(sqrt(1-p^2))/p$

$tan (90 - x) = \frac{p}{\sqrt{1 - p^{2}}}$ $tan(90-x)=p/sqrt(1-p^2)$

Explanation:

The Pythagorean theorem states

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

Solving for $sin x$ $sinx$ we have

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

$sin x = \sqrt{1 - {cos}^{2} x}$ $sinx=sqrt(1-cos^2x)$

If

$cos x = p$ $cosx=p$ ,

then

$sin x = \sqrt{1 - p^{2}}$ $sinx=sqrt(1-p^2)$ ,

$tan x = \frac{sin x}{cos x} = \frac{\sqrt{1 - p^{2}}}{p}$ $tanx=sinx/cosx=(sqrt(1-p^2))/p$ ,

and

$tan (90 - x) = \frac{sin (90 - x)}{cos (90 - x)} = \frac{cos x}{sin x} = \frac{p}{\sqrt{1 - p^{2}}}$ $tan(90-x)=sin(90-x)/cos(90-x)=cosx/sinx=p/sqrt(1-p^2)$

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Given that cosx=p Where x us an acute angle in degrees find in term of p i) Sin x ii)tanx iii)tan(90-x)?

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