Trig Precalculus - Establishing Identities

Question

Establishing Identities

cos²θ(1 + tan² θ) = 1

csc v − 1 csc v + 1 = 1 − sin v 1 + sin v

tan u − cot u / tan u + cot u + 2 cos² u = 1

cosθ + sinθ − sin³θ / sinθ = cotθ + cos³θ

Could you list out all the step I would need to verify the equation.

Answer 1

cos²θ(1 + tan² θ) = 1

tanθ=sinθ/cosθ

Thus, this equals:

cos²θ + sin² θ =1, which is the basic trig identity. (see the unit circle/pythagorean theorem for proof)

csc v − 1 csc v + 1 = 1 − sin v 1 + sin v

This one is just addition I think? Unless that v1 actually means something...

so it evaluates to 1=1

tan u − cot u / tan u + cot u + 2 cos² u = 1

cot u / tan u =(Cosu/Sinu)/(Sinu/Cosu)=(Cosu/Sinu)²

so this evaluates to

Sinu/Cosu-(Cosu/Sinu)²+Cosu/Sinu+2 cos² u = 1

(Sin³u-Cos²u+Cos²u*Sinu)/(Sin²u*Cosu)+2 cos² u =1

Okay, at this point I went online and checked the identity, and plugging in 1 for u gave me 2.37... so I'm assuming there was a typo here somewhere.

cosθ + sinθ − sin³θ / sinθ = cotθ + cos³θ

First plug in some value for theta to verify whether this is true. Plugging in 1 for theta, we obtain the left side is equivalent to circa. .67 and the right evaluates to circa .799, so the identity is false.

If you need to somehow demonstrate that the two sides are uneven (perhaps by simplifying all terms to single powers of sine/cosine), please notify me.