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)`
Solution
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
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Solution If `Sec a = 17/8` Verify that `(3 - 4sin^2a)/(4 Cos^2 a - 3) = (3 - Tan^2 A)/(1 - 3 Tan^2 A)` Concept: Trigonometric Ratios.
