How do you find the remaining trigonometric functions of $θ$ $theta$ given $tan θ = \frac{a}{b}$ $tantheta=a/b$ where a and b are both positive?

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How do you find the remaining trigonometric functions of $θ$ $theta$ given $tan θ = \frac{a}{b}$ $tantheta=a/b$ where a and b are both positive?

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Answer 1 · 2017-05-24T07:42:47

Please see below.

Explanation:

As $tan θ = \frac{a}{b}$ $tantheta=a/b$ , $cot θ = \frac{1}{\frac{a}{b}} = 1 \times \frac{b}{a} = \frac{b}{a}$ $cottheta=1/(a/b)=1xxb/a=b/a$

$sec θ = \pm \sqrt{1 + {tan}^{2} θ} = \pm \sqrt{1 + \frac{a^{2}}{b^{2}}} = \pm \frac{\sqrt{a^{2} + b^{2}}}{b}$ $sectheta=+-sqrt(1+tan^2theta)=+-sqrt(1+a^2/b^2)=+-sqrt(a^2+b^2)/b$

$cos θ = \pm \frac{b}{\sqrt{a^{2} + b^{2}}}$ $costheta=+-b/sqrt(a^2+b^2)$

$sin θ = tan θ \times cos θ = \pm \frac{a}{b} \times \frac{b}{\sqrt{a^{2} + b^{2}}} = \pm \frac{a}{\sqrt{a^{2} + b^{2}}}$ $sintheta=tanthetaxxcostheta=+-a/bxxb/sqrt(a^2+b^2)=+-a/sqrt(a^2+b^2)$

and $csc θ = \pm \frac{\sqrt{a^{2} + b^{2}}}{a}$ $csctheta=+-sqrt(a^2+b^2)/a$

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How do you find the remaining trigonometric functions of $θ$ $theta$ given $tan θ = \frac{a}{b}$ $tantheta=a/b$ where a and b are both positive?

1 Answer

Explanation:

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Science

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How do you find the remaining trigonometric functions of θtheta given tanθ=abtantheta=a/b where a and b are both positive?

1 Answer

Explanation:

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How do you find the remaining trigonometric functions of $θ$ $theta$ given $tan θ = \frac{a}{b}$ $tantheta=a/b$ where a and b are both positive?