prove that- (1) (tanθ)/(1-cotθ)+(cotθ)/(1-tanθ)=1+tanθ+cot θ (2) (tan^3A-1)/(tanA-1)=sec^2A+tanA

(1) (tanθ)/(1-cotθ)+(cotθ)/(1-tanθ)=1+tanθ+cot θ

L.H.S=(tanθ)/(1-cotθ)+(cotθ)/(1-tanθ)

=(sinθ/cosθ)/{1-(cosθ/sinθ)}+(cosθ/sinθ)/{1-(sinθ/cosθ)}

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

=sin^2 θ/(cosθ[sinθ-cosθ]) + cos^2 θ/(sinθ[cosθ-sinθ])

=sin^2 θ/(cosθ[sinθ-cosθ]) - cos^2 θ/(sinθ[sinθ-cosθ])

=1/(sinθ-cosθ){sin^2 θ/cosθ - cos^2 θ/sinθ}

=1/(sinθ-cosθ){(sin^3 θ - cos^3 θ)/sinθcosθ}

=1/(sinθ-cosθ){[(sinθ - cosθ)(sin^2 θ+sinθcosθ+cos^2 θ)]/sinθcosθ}

=(sin^2 θ+sinθcosθ+cos^2 θ)/sinθcosθ

=sin^2 θ/sinθcosθ + sinθcosθ/sinθcosθ + cos^2 θ/sinθcosθ

=sinθ/cosθ + 1 + cosθ/sinθ

=tanθ+1+cotθ

=1+tanθ+cot θ

=R.H.S

(2) (tan^3 A-1)/(tanA-1)=sec^2 A+tanA

L.H.S=(tan^3 A-1)/(tanA-1)

=([sin^3 A/cos^3 A]-1)/([sinA/cosA]-1)

=(sin^3 A-cos^3 A)/(cos^2 A[sinA-cosA])

={(sinA - cosA)(sin^2 A+sinAcosA+cos^2 A)}/(cos^2 A[sinA-cosA])

=(sin^2 A+sinAcosA+cos^2 A)/cos^2 A

=(sin^2 A+cos^2 A+ sinAcosA)/cos^2 A

=(1+sinAcosA)/cos^2 A     [because sin^2 A+cos^2 A=1]

=1/cos^2 A + sinAcosA/cos^2 A

=1/cos^2 A + sinA/cosA

=sec^2 A+tanA

=R.H.S