In a triangle ABC,
$b c o s (C + θ) + c c o s (B - θ) =$

a c o s θ

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a s i n θ

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a t a n θ

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a c o t θ

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Solution

The correct option is A $a c o s θ$
In $△ A B C, A + B + C = π$ and $a = 2 R sin A, b = 2 R sin B, c = 2 R sin C$
$b cos (C + θ) + c cos (B - θ) = R (2 sin B cos (C + θ) + 2 sin C cos (B - θ))$
$= R (sin (B + C + θ) + sin (B - C - θ) + sin (B + C - θ) + sin (C - B + θ))$
$= R (sin (π - (A - θ)) + sin (π - (A + θ)) + sin (B - C - θ) - sin (B - C - θ))$
$= R (sin (A - θ) + sin (A + θ))$
$= R (2 sin A cos θ)$
$= (2 R sin A) cos θ$
$= a cos θ$

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