Help with trig identities?

Question

i cant figure these two questions out :(

1) Express the following in terms of sin0 and/or cos0 (the 0 being beta)

1+cot^2beta/cot^2beta

2) Prove that tan^2beta/sin^2beta = 1+tan^2beta , then check the identity using beta=120 degrees

Thanks for all your help :)

Answer 1

1. Uh, seems pretty straightforward.

cot^2(beta)/cot^2(beta) = 1

1 + 1 = 2

So the answer is 2.

2.

tan^2(beta) = sin^2(beta)/cos^2(beta)

tan^2(beta)/sin^2(beta) = [sin^2(beta)/cos^2(beta)] / sin^2(beta)

tan^2(beta)/sin^2(beta) = 1/cos^2(beta)

But 1 = sin^2(beta) + cos^2(beta)

tan^2(beta)/sin^2(beta) = [sin^2(beta) + cos^2(beta)]/cos^2(beta)

tan^2(beta)/sin^2(beta) = 1 + tan^2(beta)

Use beta = 120 degrees

tan(beta) = SQRT(3) ......... tan^2(beta) = 3

sin(beta) = SQRT(3)/2 ...... sec^2(beta) = 3/4

tan^2(beta)/sin^2(beta) = 3/(3/4) = 4

1 + tan^2(beta) = 1 + 3 = 4

and 4 = 4 so it checks

Answer 2

1+cot^2beta=csc^2beta

1+cot^2beta/cot^2beta= csc^2beta / cot^2beta

=(1/sin^2beta)(sin^2beta / cos^2beta)

=1 / cos^2beta

2)tan^2beta/sin^2beta= (sin^2beta / cos^2beta)(1/sin^2beta)

=1/cos^2beta = sec^2beta = 1+tan^2beta

beta = 120 degrees

tan(120)=sqrt(3)

sin(120)=sqrt(3)/2

tan^2(120)/sin^2(120) = 3 / (3/4)=4 left side

1+tan^2(120) = 1+ sqrt(3)*sqrt(3) = 1+3 = 4 right side