है dee A= 13 *nm{«\\ thal

1.	है dee A= 13 *nm{«\\ thal
Answer» Given:secA = 17/8 We know, sec A = Hypotenuse/Base = 17/8 So, draw a right angled triangle PQR, right angled at Q such that∠ PQR = ABase QR = 8 and Hypotenuse PR = 17 Using Pythagoras Theorem in ΔPQR 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

Answer»

Given:secA = 17/8

We know, sec A = Hypotenuse/Base = 17/8

So, draw a right angled triangle PQR, right angled at Q such that∠ PQR = ABase QR = 8 and Hypotenuse PR = 17

Using Pythagoras Theorem in ΔPQR

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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