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Given 5sinA=3 find secA-tanA |
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Answer» Answer: 5sin A= 3 sinA =opp/hyp =3/5 Let BC be 3K AC be 5k (k is the positive REAL no.) By applying Pythagoras theorem AC2 = AB2 +BC2 (5)2 = (AB)2+ (3)2 25 = (AB)2+ 9 25- 9= AB2 16 = AB2 4 = AB Sec A =5/3 tan A = 3/4 secA -tanA 5/3 - 3/4 LCM of 3 & 4 =12 5 - 3/12 2 /12 6 |
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