If csc theta = √5, then what is cot theta?

- Mirandaregina

If csc theta = √5, then what is cot theta?

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- Nnesha

what is the reciprocal of csc ?

- Mirandaregina

@Nnesha Sine

- Nnesha

not just sin :=)
1/sin = csc
so csc = ?

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## More answers

- anonymous

In addition to sine, cosine, and tangent, there are three other trigonometric functions you need to know for the Math IIC: cosecant, secant, and cotangent. These functions are simply the reciprocals of sine, cosine, and tangent. Cotangent is the reciprocal of tangent.

- Mirandaregina

@Nnesha so csc = 1/sine

- Nnesha

yes right so if csc = sqrt{5} then sin = ???

- Nnesha

remember thy are reciprocal of each other

- Mirandaregina

@Nnesha then sin = 1/√5 = √5/5 ?

- phi

draw a triangle and label its opposite side 1 and its hypotenuse sqr(5)

- Nnesha

hmm \[\csc = \frac{ \sqrt{5} }{ 1 } ~~~~~ \sin=\frac{ 1 }{ \sqrt{5} }\]
so draw right triangle for now it's okay leave the radical sign at the denominator |dw:1440603990058:dw| \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

- Mirandaregina

@Nnesha so if sin theta = opp/hyp, and opp is 1 and hyp is √5, is it √5/5?

- Nnesha

yes right sin = opp .hyp|dw:1440604248339:dw|
apply the Pythagorean theorem to find 3rd side of right triangle

- phi

yes you could use sqr(5) for the opposite side and 5 for the hypotenuse
but 1 and sqr(5) seem easier (smaller numbers)
but either set will give the same answer.

- Mirandaregina

@Nnesha 1^2 + √5^2 = 6

- phi

c^2 = a^2 + b^2
the unknown is a leg.

- Nnesha

c=hypotenuse(longest leg of right triangle )

- Mirandaregina

@Nnesha @phi did I use the Pythagorean Theorem correctly? doesn't a = 1 and b = √5 ?

- phi

no a and b are the legs
you know a and c (and you need to find b, the other leg)

- phi

you know the hypotenuse (because sin is opp/hyp)
you know a side and the hypotenuse

- Nnesha

|dw:1440604696536:dw|

- Mirandaregina

@phi how do I find b?

- anonymous

quick question is 1/3 a rational number

- phi

a^2 + b^2 = c^2
put in numbers for a and c
can you do that ?

- Mirandaregina

@phi but does a = 1? because before I said 1^2 + √5^2 = 6

- phi

a=1 and c= \( \sqrt{5} \)
put those in the formula
\[a^2 + b^2 = c^2\]

- phi

notice c is on the other side of the = sign from a and b

- Nnesha

i guess she is trying to say that sqrt of 5 +1 equal to 6

- Nnesha

square of sqrt 5**

- phi

when you write
1^2 + √5^2 = c^2
a^2 + b^2 = c^2
that means a is 1 and b is sqr(5). but that is not what we want

- Mirandaregina

@phi I understand, thanks :)
so I'm not sure if I did this right but I have 1^2 + b^2 = √5
b + 1^2 = √5
b^2 + 1 - 1 = √5 - 1
b^2 = √1.236068

- phi

almost, but the formula says c^2 so you want (sqrt(5))^2

- Mirandaregina

@phi as is √1.111786 = 1.0544126

- Mirandaregina

*as in

- phi

you wrote
1^2 + b^2 = √5
but what it should be is
1^2 + b^2 = (√5)^2
now find b

- Mirandaregina

@phi oh my mistake
1^2 + b^2 = (√5)^2
b = 2

- phi

yes. now you know all 3 sides of the triangle
|dw:1440605865953:dw|

- phi

the want the cotangent of theta
(I would write the tangent of theta, then "flip it" to make it the cotangent)
or you can remember cotangent is adj/opp

- Mirandaregina

@phi cot theta = adj/opp = 2/1 = 2

- phi

yes

- Mirandaregina

@phi so that's the overall answer? I got that answer when I first tried it but I thought it was wrong! lol

- phi

now you know how to do it so you believe it

- Mirandaregina

@phi @Nnesha thank you for all your help and patience everyone :)

- Nnesha

my pleasure o^_^o

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