Question

Prove = sec theta / sec theta + tan theta - 1 + cos theta / cosec theta + cot tbeta - 1 = 1..​

Answer 1

Answer:

see the attachment you will get your answer

Answer 2

sinθ1−cosθ+tanθ1+cosθ

sinθ1−cosθ+tanθ1+cosθ = sinθ(1+cosθ)(1−cosθ)(1+cosθ)+tanθ(1−cosθ)(1+cos0)(1−cosθ)

sinθ1−cosθ+tanθ1+cosθ = sinθ(1+cosθ)(1−cosθ)(1+cosθ)+tanθ(1−cosθ)(1+cos0)(1−cosθ)=sinθ(1+cosθ)1−cos2θ+tanθ(1−cosθ)1−cos2θ

sinθ1−cosθ+tanθ1+cosθ = sinθ(1+cosθ)(1−cosθ)(1+cosθ)+tanθ(1−cosθ)(1+cos0)(1−cosθ)=sinθ(1+cosθ)1−cos2θ+tanθ(1−cosθ)1−cos2θ =sinθ(1+cosθ)sin2θ+sinθ(1−cosθ)cosθsin2θ

sinθ1−cosθ+tanθ1+cosθ = sinθ(1+cosθ)(1−cosθ)(1+cosθ)+tanθ(1−cosθ)(1+cos0)(1−cosθ)=sinθ(1+cosθ)1−cos2θ+tanθ(1−cosθ)1−cos2θ =sinθ(1+cosθ)sin2θ+sinθ(1−cosθ)cosθsin2θ = 1+cosθsin0+1−cosθcosθsinθ

sinθ1−cosθ+tanθ1+cosθ = sinθ(1+cosθ)(1−cosθ)(1+cosθ)+tanθ(1−cosθ)(1+cos0)(1−cosθ)=sinθ(1+cosθ)1−cos2θ+tanθ(1−cosθ)1−cos2θ =sinθ(1+cosθ)sin2θ+sinθ(1−cosθ)cosθsin2θ = 1+cosθsin0+1−cosθcosθsinθ = 1sinθ+cosθsinθ+1cosθsinθ−1sinθ

sinθ1−cosθ+tanθ1+cosθ = sinθ(1+cosθ)(1−cosθ)(1+cosθ)+tanθ(1−cosθ)(1+cos0)(1−cosθ)=sinθ(1+cosθ)1−cos2θ+tanθ(1−cosθ)1−cos2θ =sinθ(1+cosθ)sin2θ+sinθ(1−cosθ)cosθsin2θ = 1+cosθsin0+1−cosθcosθsinθ = 1sinθ+cosθsinθ+1cosθsinθ−1sinθ = cscθ+cot0+secθcscθ−cscθ

sinθ1−cosθ+tanθ1+cosθ = sinθ(1+cosθ)(1−cosθ)(1+cosθ)+tanθ(1−cosθ)(1+cos0)(1−cosθ)=sinθ(1+cosθ)1−cos2θ+tanθ(1−cosθ)1−cos2θ =sinθ(1+cosθ)sin2θ+sinθ(1−cosθ)cosθsin2θ = 1+cosθsin0+1−cosθcosθsinθ = 1sinθ+cosθsinθ+1cosθsinθ−1sinθ = cscθ+cot0+secθcscθ−cscθ

sinθ1−cosθ+tanθ1+cosθ = sinθ(1+cosθ)(1−cosθ)(1+cosθ)+tanθ(1−cosθ)(1+cos0)(1−cosθ)=sinθ(1+cosθ)1−cos2θ+tanθ(1−cosθ)1−cos2θ =sinθ(1+cosθ)sin2θ+sinθ(1−cosθ)cosθsin2θ = 1+cosθsin0+1−cosθcosθsinθ = 1sinθ+cosθsinθ+1cosθsinθ−1sinθ = cscθ+cot0+secθcscθ−cscθ = secθcscθ+cotθ