I cant find a book that can help me understad the fundamental indentities on precalculus.

Question

I try to look online and on my precalculus book but i still dont get it.

 

How do i simplify this?

1) simplify sec(theta)/csc(theta)

 2) simplify 1/1+cot^2(theta)

  3) simllify cot^2(theta)

 

 

 

Those someone know a book that could help me uderstand?

Answer 1

I taught precalculus from Blitzer, Ed. 2. Used copies are going for ~$10. See Amazon ISBN-10: 0131013645 and click on used. It's an older text, but the math is the same.

To solve these problems, let's use my development in:

http://www.wyzant.com/resources/answers/29169/perform_the_operation_and_simplify

 

1) simplify sec(θ)/csc(θ)

 

sec(θ)/csc(θ) = (r/x)/(r/y) = (r/x)*(y/r) = y/x = tan(θ)

2) simplify 1/(1+cot^2(θ))

[Notice how I put parentheses around the denominator. That way there is no ambiguity.]

 

1/(1+cot^2(θ)) = 1/(1+(x/y)^2)

= 1/(1+x^2/y^2); multiply top & bottom by y^2

= y^2/(y^2+x^2); replace bottom with r

 

= y^2/r^2 = (y/r)^2 = sin^2(θ)

3) simplify cot^2(θ)

Not sure what the question wants because the expression is already a single trig function.

 

cot^2(θ) = (x/y)^2

 

We know: x^2 + y^2 = r^2

 

Divide both sides by y^2

 

(x/y)^2 + 1 = (r/y)^2

 

Replace with Trig Functions:

 

cot^2(θ) + 1 = csc^2(θ)

So cot^2(θ) = csc^2(θ) - 1

Answer 2

1)

 

    SEC/ CsC Θ = ( 1/Cos Θ) /( 1/Sin Θ )

                      

                      =  SinΘ / Cos Θ = Tan Θ

 

    2 ) ( 1/ 1 + Cot ^2 Θ ) =

 

         1  / ( 1 + (Cos^2 θ/ Sin²θ ) =

    

         1 / (( Sin^2 Θ + Cos^2Θ) / Sin²Θ )

     

          1/( 1/ Sin^2 Θ ) = Sin ²Θ

 

    3 ) Cot ²Θ      / Contains only 1 term , so how much more can be simplified.

Answer 3

Hi Daisy,

              www.wolframalpha.com may be of some use….the first problem above is just tan(theta) the other two I think are in their simplest forms

 

Jim