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Mathematics 14 Online
OpenStudy (anonymous):

What value of x satisfies cot (90° − x) = 1?

OpenStudy (anonymous):

60° 135° 225° 315°

OpenStudy (anonymous):

@saifoo.khan

OpenStudy (anonymous):

@dan815

OpenStudy (anonymous):

@iGreen

OpenStudy (anonymous):

@Compassionate

OpenStudy (anonymous):

Precalc

OpenStudy (anonymous):

cotangent

OpenStudy (anonymous):

degrees

OpenStudy (welshfella):

Hint: cot = 1 / tan

OpenStudy (anonymous):

?

OpenStudy (jdoe0001):

\(\bf \textit{Cofunction Identities} \\ \quad \\ sin\left(\frac{\pi}{2}-{\color{red}{ \theta}}\right)=cos({\color{red}{ \theta}})\qquad cos\left(\frac{\pi}{2}-{\color{red}{ \theta}}\right)=sin({\color{red}{ \theta}}) \\ \quad \\ \quad \\ tan\left(\frac{\pi}{2}-{\color{red}{ \theta}}\right)=cot({\color{red}{ \theta}})\qquad cot\left(\frac{\pi}{2}-{\color{red}{ \theta}}\right)=tan({\color{red}{ \theta}}) \\ \quad \\ \quad \\ cot(90^o-{\color{red}{ x}})=1\implies tan({\color{red}{ x}})\implies tan^{-1}[tan({\color{red}{ x}})]=tan^{-1}(1) \\ \quad \\ {\color{red}{ x}}=tan^{-1}(1)\)

OpenStudy (anonymous):

135?

OpenStudy (jdoe0001):

well... .recall that tangent is really \(\bf tan(\theta)=\cfrac{opposite}{adjacent} =\cfrac{y}{x} =\cfrac{{\color{brown}{ sin}}(\theta)}{{\color{blue}{ cos}}(\theta)}\) and 135 degree angle is in the 2nd Quadrant on the 2nd quadrant "y" is positive, and "x" is negative so y/x will mean -1 for a \(\bf tan(135^o\)

OpenStudy (jdoe0001):

\(\bf tan(135^o)\) =)

OpenStudy (jdoe0001):

so... is not 135

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

225

OpenStudy (anonymous):

thx

OpenStudy (jdoe0001):

yw

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