Question

If secA = `(5)/(4)`, cerify that `(3sin"A" - 4sin^3"A")/(4cos^3"A" - 3cos"A") = (3tan"A" - tan^3"A")/(1 - 3tan^2"A")`.

Answer 1

sec A = `(5)/(4)`

⇒ cos A = `(4)/(5) = "Base"/"Hypotenuse"`
Perpendicular
= `sqrt(("Hypotenuse")^2 - ("Base")^2`
= `sqrt(25 - 16)`
= `sqrt(9)`
= 3

sin A = `"Perpendicular"/"Hypotenuse" = (3)/(5)`

tan A = `"Perpendicular"/"Base" = (3)/(4)`

To show: `(3sin"A" - 4sin^3"A")/(4cos^3"A" - 3cos"A") = (3tan"A" - tan^3"A")/(1 - 3tan^2"A")`.

L.H.S. = `(3sin"A" - 4sin^3"A")/(4cos^3"A" - 3cos"A")`

= `(3(3/5) - 4(3/5)^3)/(4(4/5)^3 - 3(4/5)`

= `(9/5 - 108/125)/(256/125 - 12/5)`

= `((225 - 108)/125)/((256 - 300)/125)`

= `(117)/(-44)`

R.H.S. = `(3tan"A" - tan^3"A")/(1 - 3tan^2"A")`

= `(3(3/4) - (3/4)^3)/(1 - 3(3/4)^2`

= `(9/4 - 27/64)/(1 - 27/16)`

= `((144 - 27)/64)/((16 - 27)/16)`

= `(117)/(64) xx (16)/(-11)`

= `(117)/(4) xx (1)/(-11)`

= `(-117)/(44)`
⇒  L.H.S. = R.H.S.