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Mathematics 10 Online
OpenStudy (anonymous):

find tan(theta) and cot(theta) if sec(theta) is 4 and sin (theta) is less than 0

OpenStudy (anonymous):

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OpenStudy (anonymous):

there is a picture of an angle whose secant is \(4\) you need the other side which you find via pyathagoras then you can take all the trig ratios you like

OpenStudy (anonymous):

do you know another way? we're using reciprocal identities, quotient identites and pythagorean identities.

OpenStudy (anonymous):

it is the same thing it all amounts to finding the third side as \[\sqrt{4^2-1^2}=\sqrt{15}\]

OpenStudy (anonymous):

sorry if did not come back im helping some one else

OpenStudy (anonymous):

you have \[\sec(\theta)=4\] so right way you know \[\cos(\theta)=\frac{1}{4}\]

OpenStudy (anonymous):

to find the other two you need \(\sin(\theta)\) which we get from the triangle as \[\sin(\theta)=-\frac{\sqrt{15}}{4}\] but you can use an identity if you like

OpenStudy (anonymous):

\[\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}\] if you compute \[-\sqrt{1-\left(\frac{1}{4}\right)^2}\] you still get \[-\frac{\sqrt{15}}{4}\] it just takes longer

OpenStudy (anonymous):

so would i use the quotient identity cot(theta)= cos(theta)/sin(theta)?

OpenStudy (anonymous):

yes but it is silly

OpenStudy (anonymous):

the denominators will cancel, might as well go right to the answer you know from the triangle

OpenStudy (anonymous):

this is the way my teacher is teaching it. first day today on this topic lol

OpenStudy (anonymous):

you get from the triangle everything just make sure you know if it is positive or negative \[\cot(\theta)=-\frac{1}{\sqrt{15}}\] for example

OpenStudy (anonymous):

i sort of get it now. still trying how to find tan(theta)

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