Question

1+cos2a+sin2a/1-cos2a+sin2a=cota

Answer 1

Proved

Explanation:

Required

Show that:

`(1+cos2a+sin2a)/(1-cos2a+sin2a)=cota`

`cos2a = cos(a + a) = cos^2a - sin^2a`

So, we have:

`(1+cos^2a - sin^2a+sin2a)/(1-(cos^2a - sin^2a)+sin2a)=cota`

`(1+cos^2a - sin^2a+sin2a)/(1-cos^2a + sin^2a+sin2a)=cota`

`1- cos^2a = sin^2a`

So, we have:

`(1+cos^2a - sin^2a+sin2a)/( sin^2a+ sin^2a+sin2a)=cota`

`(1+cos^2a - sin^2a+sin2a)/(2sin^2a+sin2a)=cota`

Rearrange the numerator

`(1 - sin^2a+cos^2a+sin2a)/(2sin^2a+sin2a)=cota`

`1- sin^2a= cos^2a`

So, we have

`(cos^2a+cos^2a+sin2a)/(2sin^2a+sin2a)=cota`

`(2cos^2a+sin2a)/(2sin^2a+sin2a)=cota`

`sin2a = 2sina\ cosa`

So, we have:

`(2cos^2a+2sinacosa)/(2sin^2a+2sinacosa)=cota`

Factorize:

`(cosa(2cosa+2sina))/(sina(2sina+2cosa))=cota`

Rewrite as:

`(cosa(2cosa+2sina))/(sina(2cosa+2sina))=cota`

`(cosa)/(sina) = cota`

`cota = cota`