if cos A 8/17, then show that 3-4sin2A/4cos2a-3=3-tan2a/1-3tan2a
every 2 is square
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Answer:
Step-by-step explanation:
cosA=8/17
then sin²A=1-cos²A
=1-(8/17)²
=1-64/289
=(289-64)/289
=225/289
Hence sinA=√225/289=15/17
============================
Now the given expression:
(3-4sin²A) / (4cos²A-3) =3-tan²A/1-3tan²A
LHS =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
LHS= RHS
proved
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