How i prove this ?

Take into account that #15º=45-30#

Appliying #sin(a-b)=sinacosb-cosasinb# and #cos(a-b)=cosacosb+senasenb#

We have #sin15=sin(45-30)=sin45cos30-cos45 sen30=sqrt2/2·sqrt3/2-sqrt2/2·1/2=sqrt6/4-sqrt2/4#

By the other hand..#cos15=cos(45-30)=cos45cos30+sin45sin30=sqrt2/2·sqrt3/2+sqrt2/2·1/2=sqrt6/4+sqrt2/4#

Finally we have #sin15+tan30cos15=sqrt6/4-sqrt2/2+sqrt3/3(sqrt6/4+sqrt2/4)=sqrt6/4-sqrt2/4+sqrt18/12+sqrt6/12=3sqrt6/12-3sqrt2/12+sqrt18/12+sqrt6/12=1/12(3sqrt6-cancel(3sqrt2)+cancel(3sqrt2)+sqrt6)=4sqrt6/12=sqrt6/3#

#sin 15 + tan 30 cos 15#

#sin 15+ (sin 30 / cos 30 ) * cos 15, " as " tan theta = (sin theta / cos theta)#

#=> (cos 30 sin 15 + sin 30 cos 15) / cos 30, " taking cos 30 as L C M"#

#sin A cos B + cos A sin B = sin (A + B)#

#:. =>sin (30 + 15) / cos 30#

#=> sin 45 / cos 30 #

We know #sin 45 = 1/ sqrt2, cos 30 = sqrt 3 / 2#

#"Hence " => (1/sqrt2) / (sqrt3 / 2) =2 / (sqrt 2 * sqrt 3)#

#=> cancel(2)^color(red)(sqrt2) / (cancelsqrt 2 * sqrt3 ) * (sqrt 3 / sqrt3), " multiply and divide by " sqrt3#

#=>(sqrt 2 * sqrt 3) / 3 = sqrt6 / 3#

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