Question

 if `sec A = 5/4` verify that `(3 sin A - 4 sin^3 A)/(4 cos^3 A - 3 cos A) = (3 tan A - tan^3 A)/(1- 3 tan^2 A)`

Answer 1

We have

`sec A = 5/4`

In ΔABC

`AC^2 = AB^2 + BC^2`

`=> (5)^2 = (4)^2 + (BC)^2`

`=> BC^2 = 25 - 16`

=> BC = 3

`sin A = 3/5, cos A = 4/5 and tan A = 3/4`

Now

`(3sin A - 4sin^3 A)/(4 cos^3 A - 3 cosA) = (3 tan A - tan^3 A)/(1 - 3tan^2 A)`

`=> (3 xx 3/5 - 4 xx (3/5)^2)/(4 xx (4/5)^3 - 3 xx (4/5)) = (3xx (3/4) - (3/4)^3)/(1 - 3xx(3/4)^2)`

`=> (9/5 - 108/125)/(256/25 - 12/5) = (9/4 - 27/64)/(1 - 27/16)`

`=> ((225 - 108)/125)/((256 - 300)/125) = ((144 - 27)/64)/((16 - 27)/16)`

`=> 117/(-44) = 117/(-11xx4)`

=> L.H.S  = R.H.S