How would I solve the following trig problems?

Question

a) 1-cos^2(-θ)/1+tan^2(-θ) 

b) (1-cos θ)(1+sec θ) 

c) (sin θ- cos θ)(csc θ + sec θ)

Update : Write each expression in terms of sine and cosine.

Answer 1

a) [1 - cos²(-θ)] / [1 + tan²(-θ)]

= {1 - [ cos(-θ)]² }/ {1 + [tan(-θ)]² }

= {1 - [ cos(θ)]²}/{1 + [ - tan(θ)]²}

= [ 1 - cos²(θ)]/[1 + tan²(θ)]

{Recall the Pythagorean identities sin²(x) + cos²(x) = 1 and 1 + tan²(x) = sec²(x)}

= sin²(θ)/sec²(θ)

= sin²(θ) cos²(θ)

Answer 2

b) ( 1 - cosθ)( 1 + secθ)

= (1 - cosθ)(1 + 1/cosθ)

= (1 - cosθ)(cosθ + 1)/cosθ

= [(1 - cosθ)(1 + cosθ)]/cosθ

{Apply the formula (a² - b² ) = (a - b)(a + b), in this case a = 1, b = cosθ }

= (1 - cos²θ)/cosθ

{From Pythagorean identity sin²(x) + cos²(x) = 1}

= (sin²θ)/(cosθ)

Answer 3

c) (sinθ - cosθ)(cscθ + secθ)

= (sinθ - cosθ)(1/sinθ + 1/cosθ)

= (sinθ - cosθ)(cosθ + sinθ)/sinθcosθ

= [(sinθ - cosθ)(sinθ + cosθ)]/sinθcosθ

{Apply the formula (a² - b² ) = (a - b)(a + b), in this case a = sinθ, b = cosθ }

= [ sin²θ - cos²θ]/sinθcosθ