Question

If cos theta = b/√a^2+b^2 then prove that

( b+√a^2+b^2/a )=√a^2+b^2 +b/√a^2+ b^2-b = 1+ cos theta/1-cos theta

Answer 1

ANSWER.................


As cosθ=cosα−e1−ecosα

cosθ−ecosθcosα=cosα−e

or e(cosθcosα−1)=cosθ−cosα

or 1e=cosθcosα−1cosθ−cosα

an using componendo dividendo

1+e1−e=cosθcosα−1+cosθ−cosαcosθcosα−1−cosθ+cosα

= cosθ(1+cosα)−(1+cosα)cosα(1+cosθ)−(1+cosθ)

= (cosθ−1)(1+cosα)(cosα−1)(1+cosθ)

= 1+cosα1−cosα×1−cosθ1+cosθ

Observe that tan2A=2sin2A2cos2A

= 1−1+2sin2A1+2cos2A−1=1−(1−2sin2A)1+(2cos2A−1)

= 1−cos2A1+cos2A

Hence 1+e1−e=tan2(θ2)tan2(α2)

and tan(θ2)=±√1+e1−etan(α2)

$= = = = = = = = = = = = =$

Answer 2

Step-by-step explanation:

from abhijeet jamui bihar