Question

FInd exact value of: csc (37) sec(53) - tan (53) cot (37)

Answer 1

http://ichthyosapiens.com/School/Math/Trig/ch05CofunctionIdentities.jpg

can you rewrite the csc(37) you think?

Answer 2

keep in mind that 90-53 = 37

Answer 3

Ok, I'll try, hold on...

Answer 4

1/sin (37) * 1/cos (37) - sin (37)/cos (37) * cos (37)/sin (37)
Is this right?

Answer 5

hmm did you see the "cofunction identities" ?

Answer 6

Oh, ok I'll look...

Answer 7

1/sin (37) * 1/sin (37) - cos (37)/sin (37) * cos (37)/sin (37)

Answer 8

hmm gimme one sec

Answer 9

Ok.

Answer 10

so using the cofunction identities -> http://ichthyosapiens.com/School/Math/Trig/ch05CofunctionIdentities.jpg then we get \( \large \begin{array}{llll}
csc (37^o) &sec(53^o) - tan (53^o)& cot (37^o)\\ \quad \\
csc(90^o-{\color{red}{ 53}}^o)&sec(53^o) -tan (53^o)&cot(90^o-{\color{red}{ 53}}^o)\\ \quad \\
sec({\color{red}{ 53}}^o)&sec(53^o) -tan (53^o)&tan({\color{red}{ 53}}^o)
\end{array}\)

Answer 11

so... what would that give you anyway?

Answer 12

I thought I was supposed to put everything in terms of sin and cos, then simply...?

Answer 13

well... yes, I gather you could do it that way too

Answer 14

So does that make my second answer (without the simplication) correct?

Answer 15

Or did I do that wrong?

Answer 16

nope, looks good the way you have it

Answer 17

using the cofunctions identities you'd end up with -> 1/sin (37) * 1/sin (37) - cos (37)/sin (37) * cos (37)/sin (37)

Answer 18

1/sin (37) * 1/sin (37) - cos (37)/sin (37) * cos (37)/sin (37)
Simplified:

1/sin (37) - 2 cos (37)/sin (37) ?

Not sure how to simplify this further...?

Answer 19

Wouldn't that give me a negative?

-cos (37)/sin (37)

Answer 20

\(\bf \left[\cfrac{1}{sin(37^o)}\cdot \cfrac{1}{sin(37^o)}\right]-\left[\cfrac{cos(37^o)}{sin(37^o)}\cdot \cfrac{cos(37^o)}{sin(37^o)}\right]\\ \quad \\
\implies \cfrac{1^2}{sin^2(37^o)}-\cfrac{cos^2(37^o)}{sin^2(37^o)}\)

Answer 21

or -tan (37) ?

Answer 22

well, not quite

Answer 23

if we add those fractions, we get \(\bf \cfrac{1^2}{sin^2(37^o)}-\cfrac{cos^2(37^o)}{sin^2(37^o)}\implies \cfrac{{\color{blue}{ 1-cos^2(37^o)}}}{sin^2(37^o)}\\ \quad \\
recall\implies sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)={\color{blue}{ 1-cos^2(\theta)}}\)

Answer 24

I don't see how you got 1- cos^2 (37)...?

Answer 25

well notice the LCD from the two fractions

Answer 26

both fractions have the same denominator, thus the LCD will be just that, then the numerator will simply be itself moved over

Answer 27

\(\bf 1^2=1\)

Answer 28

Thanks.

Answer 29

yw