Question

I really need help verifying this proof (Trignometric Proof)

Answer 1

\[LS\]\[=\frac{tan\theta}{1-cot \theta}+\frac{cot \theta}{1-tan\theta}\]\[=\frac{tan\theta}{1-\frac{1}{tan\theta}}+\frac{cot \theta}{1-tan\theta}\]\[=\frac{tan\theta}{\frac{tan\theta -1}{tan\theta}}+\frac{cot \theta}{1-tan\theta}\]\[=\frac{tan^2\theta}{tan\theta -1}+\frac{cot \theta}{1-tan\theta}\]\[=-\frac{tan^2\theta}{1-tan\theta}+\frac{cot \theta}{1-tan\theta}\]\[=\frac{cot \theta-tan^2\theta }{1-tan\theta}\]\[=\frac{1-tan^3\theta }{ tan \theta (1-tan\theta)}\]\[=\frac{(1-tan \theta)(tan^2 \theta+tan\theta+1)}{ tan \theta (1-tan\theta)}\]\[=\frac{(tan^2 \theta+tan\theta+1)}{ tan \theta }\]\[=(tan\theta+\frac{1}{tan\theta}+1)\]\[=...\]

Answer 2

using tanθ=sinθ/cosθ and cotθ=cosθ/sinθ
we have
= tanθ/(1-cotθ) + cotθ/(1-tanθ)

Answer 3

thus we
=sinθ*sinθ/(cosθ(sinθ-cosθ)) + cosθ*cosθ/(sinθ(cosθ-sinθ))

Answer 4

= sin^2θ/(cosθ(sinθ-cosθ)) -cos^2θ/(sinθ(sinθ-cosθ))
= (sin^3θ-cos^3θ)/(cosθ*sinθ*(sinθ-cosθ))
=(sinθ-cosθ)(sin^θ+sinθcosθ+cos^2θ)/(cosθ*sinθ*(sinθ-cosθ))

Answer 5

=(sin^θ+sinθcosθ+cos^2θ)/cosθ*sinθ =(1+cosθ*sinθ)/cosθ*sinθ
=1/cosθ*sinθ +cosθ*sinθ/cosθ*sinθ=secθcosecθ +1

Answer 6

using tanθ=sinθ/cosθ and cotθ=cosθ/sinθ we have
LHS = tanθ/(1-cotθ) + cotθ/(1-tanθ)
=sinθ*sinθ/(cosθ(sinθ-cosθ)) + cosθ*cosθ/(sinθ(cosθ-sinθ))
= sin^2θ/(cosθ(sinθ-cosθ)) -cos^2θ/(sinθ(sinθ-cosθ))
= (sin^3θ-cos^3θ)/(cosθ*sinθ*(sinθ-cosθ))
=(sinθ-cosθ)(sin^θ+sinθcosθ+cos^2θ)/(cosθ*sinθ*(sinθ-cosθ))
=(sin^θ+sinθcosθ+cos^2θ)/cosθ*sinθ
=(1+cosθ*sinθ)/cosθ*sinθ
=1/cosθ*sinθ +cosθ*sinθ/cosθ*sinθ
=secθcosecθ +1
=RHS