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Question

If sinA:cosA=4:7, then the value of 7sinA−3cosA7sinA+2cosA is

A
314
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B
32
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C
13
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D
16
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Solution

The correct option is D 16
sinAcosA=47

S=7sinA3cosA7sinA+2cosA
Divide numerator and denominator by cosA, we get
S=7sinAcosA37sinAcosA+2=434+2=16

Hence, option D.

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