Thread: Trigonometry Identities?

Originally Posted by brewerangel

Prove each of the following identities algebraically:

a) cos(x) + (cos(x)*tan(x)^2 = sec(x)

b) cos(x)(sec(x)-csc(x))=1-cot(x)

Hello,

to a)

Code:

                                     cos(x) * (sin(x))²
cos(x) + cos(x)*(tan(x))² = cos(x) + ------------------- =
                                        (cos(x))²

         (sin(x))²    cos(x)*cos(x) + (sin(x))²
cos(x) + --------- = ------------------------- = 
          cos(x)               cos(x)

     1
 ---------- = sec(x)
   cos(x)

Originally Posted by brewerangel

Prove each of the following identities algebraically:

a) cos(x) + (cos(x)*tan(x)^2 = sec(x)

cos(x) + cos(x) tan(x)^2 = cos(x) + cos(x) sin^2(x)/cos^2(x)

................. = cos(x) + sin^2(x)/cos(x) = [cos^2(x) + sin^2(x)]/cos(x)

................. = 1/cos(x) = sec(x)

b) cos(x)(sec(x)-csc(x)) = 1-cot(x)

cos(x)(sec(x)-csc(x)) = cos(x) [1/cos(x) - 1/sin(x)]

.................... = cos(x) [ (sin(x) - cos(x))/(cos(x) sin(x) ]

.................... = (sin(x) - cos(x))/ sin(x)

.................... = 1 - cos(x)/sin(x) = 1- cot(x)

RonL

Originally Posted by brewerangel

Prove each of the following identities algebraically:

a) cos(x) + (cos(x)*tan(x)^2 = sec(x)

b) cos(x)(sec(x)-csc(x))=1-cot(x)

Hello,

to b)

Code:

                                    1         1
cos(x)(sec(x) - csc(x)) = cos(x)*(------- - -------) =
                                   cos(x)   sin(x)

cos(x)   cos(x)
------ - ------ = 1 - cot(x)
cos(x)   sin(x)

Hello, brewerangel!

Another approach . . .

a) .cos(x) + cos(x)·tan²(x) .= .sec(x)

The left side is: .cos(x) + cos(x)·tan²(x)

Factor: . . . . . . .cos(x)·[1 + tan²(x)]

Then we have: . cos(x)·sec²(x) .= .cos(x)·sec(x)·sec(x)


Since cos(x)·sec(x) = 1, we have: .sec(x)

b) .cos(x)[sec(x)-csc(x)] .= .1 - cot(x)

The left side is: .cos(x)·sec(x) - cos(x)·csc(x)

. . . . . . . . . . . . . . cos(x)
And we have: . .1 - -------- . = . 1 - cot(x)
. . . . . . . . . . . . . . sin(x)