If seca=17/8 then prove that 3-4sin²a/4cos²a-3=3-tan²a/1‐3tan²a
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Solution:-
There is a mistake in this question. Question should be like this:-
secA = 17/8, show that 3-4sin^2A/4cos^2A-3 = 3tan^2A/1-3tan^2A.
Given sec A = 17/8
We know, sec A = Hypotenuse/Base = 17/8
So, draw a right angled triangle PQR, right angled at Q such that ∠ PQR = A
Base QR = 8 and Hypotenuse PR = 17
Using Pythagoras Theorem in Δ PQR
PR² = PQ² + QR²
⇒ (17)² = PQ² + (8)²
⇒ 289 = PQ² + 64
⇒ PQ² = 289 - 64
⇒ PQ² = 225
⇒ PQ = 15
L.H.S. = 3-4sin²A/4cos²A-3
⇒ 3-4(15/17)²/4(8/17)² - 3
⇒ 3-4×(225/289)/4×(64/289) - 3
⇒ {(867-900)/289}/{(256-867)/289}
⇒ -33/289 × 289/-611
= 33/611
R.H.S = (3-tan²A)/(1-3tan²A)
⇒ 3-(15/8)²/1-3(15/8)²
⇒ 3 - (255/64)/1 - (675/64)
⇒ -33/64 × 64/-611
⇒ 33/611
So, L.H.S. = R.H.S.
Hence proved
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