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Question

If a tan θ = b, then a cos 2θ + b sin 2θ =

[EAMCET 1981, 82; MP PET 1996; J & K 2005]


A

a

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B

b

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C

-a

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D

-b

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Solution

The correct option is B

a


Given that tan θ = ba.

Now, a cos 2θ + b sin 2 θ = a(1tan2θ1+tan2θ) + b(2tanθ1+tan2θ)

Putting tanθ = ba, we get

= a1b2a21+b2a2 + b2ba1+b2a2 = a(a2b2a2+b2) + b(2baa2+b2)

= 1(a2+b2) a3ab2+2ab2 = a(a2+b2)a2+b2 = a.


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