Question

Prove this Precalculus 2

Answer 1

To review (or remember):

tanФ=sinФ/cosФ (a)
cotФ = cosФ/sinФ (b)
sin2Ф = 2sinФ.cosФ (c)
and (1/2).sin2Ф = sinФ.cosФ

SOLVE: [sin²Ф-tanФ]/[cos²Ф-cotФ] = tan²Ф

1) Numerator could be written as: [sin²Ф - (sinФ/cosФ)]
OR [sin²Ф.cosФ - sinФ]/cosФ

2) Denominator could be written as: [cos²Ф - (cosФ/sinФ)]
OR [sin.cos²Ф - cosФ]/sinФ

N/D =  {[sin²Ф.cosФ - sinФ]/cosФ} ÷{ [sin.cos²Ф - cosФ]/sinФ}

(remember: (a/b)÷(c/d) = (a/b) x (d/c). We have to apply and at the same time put in the numerator sinФ into factor and in the denominator cosФ into factor.

So: (cosФ.sin²Ф - sinФ) / (cos²Ф.sinФ - cosФ) x (sinФ/cosФ)
Now put again sinФ as a factor in the numerator and cosФ in the denominator

→→{[sinФ(cosФ.sinФ - 1] / [cosФ(cosФ.sinФ - 1]} x (sinФ/cosФ)

Simplify by (cosФ.sinФ - 1)

→→ (sinФ/cosФ) x (sinФ/cosФ) = (tanФ) x (tanФ) = tan²Ф