Question

if cosec A = 17/8,tgen verify that 3-4sin^2A/4cos^2a-3=3-tan^2A/1-3tan^2A THE FIRST ANSWER WILL BE BRAINLIEST​

Answer 1

Step-by-step explanation:

show that =

4cos

2

A−3

3−4sin

2

A

=

1−3tan

2

A.

3tan

2

A

Given sec A =

8

17

We know, sec A =

Base

Hypotenuse

=

8

17

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