If the curved surface area , volume height and semi- vertical angle of a right circular cone be S,V ,h and `alpha` respectively , then prove that `S=(pi h^2sina)/(cos^2a) and V=(1)/(3)pi h^3tan^3alpha`.

Explanation

Here , the height , `AB=h, angle BAC = alpha =` semi - vertical angle , radius of base , `BC=r` and slant height, AC =l.
`therefore (r)/(h)=tan alpha or , h tan alpha = r` ...................(1)
and `(h)/(l)= cos alpha or , l =(h)/(cos alpha)` ...............(2)
Now , the curved surface area of the cone `=pi rl` .
`= pi xx h tan alpha xx(h)/(cos alpha) ` [by (1) and (2)] `=pi xx h.(sinx)/(cos a)xx(h)/(cos alpha)=(pi h^2 sin alpha)/(cos^2alpha)`
`therefore S=(pi h^2sin alpha)/(cos^2alpha)` [Proved]
Moreover , the volume of the cone , `V=(1)/(3)pi r^2h=(1)/(3) pi (h tan alpha)^2.h` [by (1)].
`=(1)/(3)pi .h^2tan^2alpha.h=(1)/(3)pi h^3tan^2alpha`
Hence `V=(1)/(3)pi h^3tan^2alpha`.