Question

`(1+sinx -cosx) /(1+cosx +sinx )=tan(x/2)` prove the identities?help

Answer 1

#LHS=(1+sinx -cosx )/(1+cosx +sinx ) #

#=(sinx(1+sinx -cosx ))/(sinx(1+cosx) +sin^2x ) #

`=(sinx(1+sinx -cosx ))/(sinx(1+cosx) +(1-cos^2x) )`

`=(sinx(1+sinx -cosx ))/((1+cosx)(sinx +1-cosx) )`

`=sinx/(1+cosx)`

`=(2sin(x/2)cos(x/2))/(2 cos^2(x/2))`

`=sin(x/2)/cos(x/2)`

`=tan(x/2)`

Answer 2

See a Proof in Explanation.

Explanation:

We have, `(color(red)(1)+color(blue)(sinx))/color(brown)cosx`,

`={color(red)(cos^2(x/2)+sin^2(x/2))+color(blue)(2cos(x/2)sin(x/2))}/{color(brown)(cos^2(x/2)-sin^2(x/2))}`,

`={cos(x/2)+sin(x/2)}^cancel(2)/[cancel({cos(x/2)+sin(x/2)}){cos(x/2)-sin(x/2)}]`.

`rArr (1+sinx)/cosx={cos(x/2)+sin(x/2)}/{cos(x/2)-sin(x/2)}.`

By dividendo-componendo

`{(1+sinx)-cosx}/{(1+sinx)+cosx}`,

`=[{cos(x/2)+sin(x/2)}-{cos(x/2)-sin(x/2)}]/[{cos(x/2)+sin(x/2)}+{cos(x/2)-sin(x/2)}]`,

`=sin(x/2)/(cos(x/2))`

`=tan(x/2),` as Respected dk_ch has readily obtained!

Enjoy Maths.!