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Anonymous asked in Education & ReferenceHomework Help · 6 years ago

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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  • 6 years ago
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    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..

  • 6 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.

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