Question

if `sec A = 17/8` verify that `(3 - 4sin^2A)/(4 cos^2 A - 3) = (3 - tan^2 A)/(1 - 3 tan^2 A)`

Answer 1

We know `sec A = "ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒"/"𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒"`

Consider right-angled triangle ABC

Let x be the adjacent side

By applying Pythagoras we get

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²

(17)² = 𝑥² + 64

𝑥² = 289 − 64

𝑥² = 225 ⇒ 𝑥 = 15

`sin A = (AB)/(BC) = 15/17`

`cos A = (BC)/(AC) = 8/17`

`tan A = (AB)/(BC) = 15/8`

`L.H.S = (3 - 4sin^2A)/(4 cos^2 A - 3) = (3 - 4 xx (15/17)^2)/(4xx (8/17)^2 - 3) = (3 -4 xx 225/289)/(4 xx 64/289 - 3) = (867 - 900)/(256 - 867) = (-33)/(-611) = 33/611`

`R.H.S = (3 - tan^2A)/(1 - 3tan^2 A) = (3 - (15/8)^2)/(1- 3 xx (15/8)^2) = (3 - (225)/64)/(1- 3 xx (225)/64) = ((-33)/64)/((-611)/64) = (-33)/(-611) = 33/611`

∴ LHS = RHS