# Trigonometric Identities: Please help!! Assignment due very very soon!!?

Prove the following identities:

a) sinθ/1+cosθ + 1+cosθ/sinθ = 2 cosec θ

b) cos^4 θ-sin^4 θ + 1=2cos^2 θ

Please show me how you work it out as i have exams in two weeks and i really am willing to learn!!!

Thanks sooo much!!

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### 2 Answers

- 6 years agoFavorite Answer
a) sinθ/(1+cosθ) + (1+cosθ)/sinθ = 2 cosec θ

Consider LHS;

sinθ/(1+cosθ) + (1+cosθ)/sinθ

(sin^2 θ + (1+cos θ)^2)/((1+cos θ)(sin θ)) [since (x+y)^2=x^2 + y^2 +2xy applying it to (1+cos θ)^2]

( sin^2 θ + (1+cos^2 θ+2cos θ) ) / ((1+cos θ)(sin θ)) [substituting sin^2 θ + cos^2 θ=1]

(2+2cos θ)/ ((1+cos θ)(sin θ))

2(1+ cos θ ) / ((1+cos θ)(sin θ)) [cancelling terms in numerator and denominator]

2/sinθ [since 1/sinθ=cosecθ]

2 cosecθ = RHS

Hence proved

b.)

we know that a^2 - b^2 =(a+b)(a-b)

Similarly we can write considering a=cos^2 θ and b=sin^2 θ

cos^4 θ-sin^4 θ + 1= (cos^2 θ + sin^2 θ)(cos^2 θ - sin^2 θ ) + 1 [substituting sin^2 θ + cos^2 θ=1]

=(1)(cos^2 θ - sin^2 θ ) +1

[we know,sin^2 θ + cos^2 θ=1 that implies sin^2 θ=1- cos^2 θ]

substitute sin^2 θ=1- cos^2 θ

= cos^2 θ - (1-cos^2 θ) +1

= 2cos^2 θ

=RHS

Hence proved..

- ChocolateHugsLv 46 years ago
a) Taking lcm,

= (sin^2 θ + (1+cosθ)^2) / sinθ (1+cosθ)

= (sin^2 θ + 1 + cos^2 θ + 2cos θ) / sin θ (1 + cos θ)

= (sin^2 θ + 1 + (1-sin^2 θ) + 2cos θ ) / sin θ(1 + cos θ)

= (sin^2 θ + 2 -sin^2 θ + 2cos θ ) / sin θ(1 + cos θ)

= (2 + 2 cos θ) / sin θ(1 + cos θ)

= { 2(1 + cos θ) } /sin θ (1 + cos θ)

= 2/sin θ

= 2 cosec θ

= RHS

b) As for this question, I couldn't prove it.