Question

sec A =17/8,verify that 3-4sin^2A/4cos^A-3=3-tan^A/1-3tan^A

Answer 1

➡HERE IS YOUR ANSWER⬇

Given that :

secA = 17/8

or, 1/cosA = 17/8

So, cosA = 8/17 => (cosA)^2 = 64/289

Then, sinA = 15/17 => (sinA)^2 = 225/289

Hence, tanA = 15/8 => (tanA)^2 = 225/64

Now,

L.H.S.

= {3 - 4 (sinA)^2} / {4 (cosA)^2 - 3}

= {3 - 4×(225/289)} / {4×(64/289) - 3}

= (3 - 900/289) / (256/289 - 3)

= (-33/289) / (-611/289)

= 33/611

R.H.S.

= {3 - (tanA)^2} / {1 - 3 (tanA)^2}

= (3 - 225/64) / {1 - 3×(225/64)}

= (3 - 225/64) / (1 - 675/64)

= (-33/64) / (-611/64)

= 33/611

Therefore, L.H.S. = R.H.S. (Proved)

⬆HOPE THIS HELPS YOU⬅

Answer 2

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