Prove each of the following identities

Question

a) cos^2 θ - sin^2 θ / cos^2 θ + sinθcosθ = 1 - tanθ

b) (sinθ + cosθ) (1+tan^2 θ) / (tanθ) = secθ + cscθ

c) 1 + tanθ / 1 + cotθ = 1 - tanθ / cotθ - 1 

d) sin^2 θ - sin^6 θ + cos^2 θ - cos^6 θ = 3sin^2 θcos^2 θ

 

Answer 1

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

L.H.S =>  (cos^2 θ - sin^2 θ)/{cosθ(cosθ + sinθ)}

=> {(cosθ + sinθ)(cosθ - sinθ)} / {cosθ(cosθ + sinθ)}

=> (cosθ - sinθ)/cosθ

=> 1 - tanθ = R.H.S

b) (sinθ + cosθ) (1+tan^2 θ) / (tanθ) = secθ + cscθ

L.H.S => (sinθ + cosθ) (1+tan^2 θ) /(tanθ)

=> (sinθ + cosθ) (sec^2{θ}) /(tanθ)

=> (sinθ + cosθ) (1/cos^2{θ}) /(tanθ)

=>  (sinθ + cosθ) {1/(sinθcosθ) }

=> sinθ/(sinθcosθ) + cosθ/(sinθcosθ)

=> secθ + cscθ = R.H.S

c) (1 + tanθ) / (1 + cotθ) = (1 - tanθ) / (cotθ - 1)

L.H.S =>  (1 + tanθ) / (1 + cotθ)

=> {1 + sinθ/cosθ}/{1 + cosθ/sinθ}

=> {(cosθ + sinθ)/cosθ}/{(sinθ + cosθ)/sinθ

=> sinθ/cosθ

=> tanθ

R.H.S => (1 - tanθ) / (cotθ - 1)

=> {1 - sinθ/cosθ}/{cosθ/sinθ  -  1}

=> {(cosθ - sinθ)/cosθ}/{(cosθ - sinθ)/sinθ}

=> sinθ/cosθ

=> tanθ

Hence, L.H.S = R.H.S

d) sin^2 θ - sin^6 θ + cos^2 θ - cos^6 θ = 3sin^2 θcos^2 θ

rearranging terms in L.H.S

=> sin^2 θ  + cos^2 θ - sin^6 θ - cos^6 θ

=> 1 - (sin^6 θ - cos^6 θ)

=> 1 - { (sin^2 θ)^3  + (cos^2 θ)^3}

Applying a^3 + b^3 = (a+b)^3 − 3ab(a+b)]

=> 1 - { (sin^2 θ + cos^2 θ)^3 - 3sin^2 θ * cos^2 θ(sin^2 θ+ cos^2 θ)}

=> 1 - { 1  - 3sin^2 θ * cos^2 θ(1)}

=> 3sin^2 θ * cos^2 θ = R.H.S