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Question

If tanθ=ab, then the value of E =asinθbcosθasinθ+bcosθ is


A

(a2 – b2)/(a2 + b2)

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B

(a-b)/(a+b)

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C

(a+b)/(a-b)

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D

(a-b)/(a2 + b2)

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E

(a2 + b2)/(a2 - b2)

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Solution

The correct option is A

(a2 – b2)/(a2 + b2)


Dividing both the numerator and the denominator of E by cos θ, we have :

E=atanθbatanθ+b=a(ab)ba(ab)+b=a2b2a2+b2


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