Question

If secA = `(17)/(8)`, verify that `(3 - 4sin^2 "A")/(4 cos^2 "A" - 3)= (3 - tan^2"A")/(1 - 3tan^2"A")`

Answer 1

secA = `(17)/(8)`

⇒ cos A = `(8)/(17) = "Base"/"Hypotenuse"`
Perpendicular
= `sqrt(("Hypotenuse")^2 - ("Base")^2`
= `sqrt((17)^2 - (8)^2`
= `sqrt(289 - 64)`
= `sqrt(225)`
= 5

sin A = `"Perpendicular"/"Hypotenuse" = (15)/(17)`

tan A = `"Perpendicular"/"Base" = (15)/(8)`

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

L.H.S. = `(3 - 4sin^2 "A")/(4 cos^2 "A" - 3)`

= `(3 - 4(15/17)^2)/(4(8/17)^2 - 3)`

= `(3 - 900/289)/(256/289 - 3)`

= `((867 - 900)/289)/((256 - 867)/289)`

= `(-33)/(-611)`

= `(33)/(611)`

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

= `(3 - (15/8)^2)/(1 -3(15/8)^2)`

= `(3 - 225/64)/(1 - 675/64)`

= `((192 - 225)/64)/((64 - 675)/64)`

= `(-33)/(-611)`

= `(33)/(611)`.