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Answer 1

Hi there

here is your answer

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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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Answer 2

Answer:

Brainliest solution

Step-by-step explanation:

LHS=RHS=33/611