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serg [7]
3 years ago
14

If 2tanA=3tanB,then prove that tan(A-B)=sin2B/(5-cos2B) ...?

Mathematics
1 answer:
andreev551 [17]3 years ago
8 0
Tan ( A - B ) = ( tan A - tan B ) / ( 1 + tan A tan B )
tan A = 3 tan B/2
tan ( A - B ) = ((3 tan B/ 2)-tan B) / ( 1 + 3 tan² B/2)=
= (tan B/2)  / ( 2 + 3 sin²B/cos²B )=
= (sin B / cos B) / (( 2cos² B+3sin²B)/cos²B)=
=( sin B cos B ) / ( 2 cos²B + 3 ( 1 - cos² B ) ) =
= (sin B cos B ) / ( 2 cos² B + 3 - 3 cos² B ) =
= ( sin 2 B ) / 2 ( 3 - cos² B ) =
= ( sin 2 B ) / ( 6 - cos² B )=
= ( sin 2 B ) / ( 5 + 1 - 2 cos² B )=
= ( sin 2 B ) / ( 5 + sin² B + cos ² B - 2 cos² B ) =
= ( sin 2 B ) / ( 5 - ( cos² B - sin² B ) ) =
= ( sin 2 B ) / ( 5 - cos 2 B )  - correct 
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Answer:

-\frac{2}{3}i - \frac{13}{5}j

Step-by-step explanation:

-0.4(3i + 4i) - 0.3(-2i + 5j) + 0.2(-\frac{1}{3}i + \frac{5}{2}j)

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Converting to fraction form;

-\frac{6}{5}i + \frac{3}{5}i - \frac{1}{15}i - \frac{8}{5}j - \frac{3}{2}j + \frac{1}{2} j

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<u>Solving the j part;</u>

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Step-by-step explanation:

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