Show `sinthetatanthetagt2(1-costheta)` if `0<theta<pi/2` :

You cannot assume that the inequality holds, so you cannot work across the inequality. (e.g. you cannot add to both sides, etc... This is assuming that the inequality is valid.)

`sinthetatantheta=sintheta(sintheta)/(costheta)=(sin^2theta)/costheta` Use the Pythagoren relationship

`=(1-cos^2theta)/costheta=((1+costheta)(1-costheta))/costheta`

`=(1+costheta)/costheta (1-costheta)`

Consider `(1+costheta)/costheta=1/costheta+1`

On `0<theta<pi/2` we have `0<costheta<1 ==> 0<1<1/costheta`

Thus `1/costheta+1>1+1=2`

Therefore `sinthetatantheta=(1/costheta+1)(1-costheta)>2(1-costheta)` as required.

`sin(theta)tan(theta)>2(1-cos(theta))`

`sin(theta){sin(theta)/cos(theta)}>2(1-cos(theta))`

since

`0<theta<pi/2`

`therefore`

`cos(theta),sin(theta),tan(theta)>0`

`sin^2(theta)>2cos(theta)(1-cos(theta))`

`sin^2(theta)+cos^2(theta)>2cos(theta)-cos^2(theta)`

`1+cos^2(theta)-2cos(theta)>0`

`(1-cos(theta))^2>0`

Which is always true. Thus

`sin(theta)tan(theta)>2(1-cos(theta))` **is true.**