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How would I solve the following trig problems?

0 votes
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.
asked Nov 6, 2014 in TRIGONOMETRY by anonymous

3 Answers

0 votes

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)

answered Nov 6, 2014 by david Expert
0 votes

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)

answered Nov 6, 2014 by david Expert
0 votes

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

answered Nov 6, 2014 by david Expert

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