If cos(A-B)=35andtanA tanB=2 then which one of the following is true ?

cos(A-B)=35 tan A tan B=2 sin A sin B=2cos A cos B cos (A-B)=35 cos A cos B+sin A sin B=35 cos A cos B+2cos A cos B=35 3cos A cos B=35 ⇒ cos A cos B=15 sin A sin B=25 cos(A+B)=cos A cos B-sin A sin B =15-25 =-15

If cos(A-B)=35andtanA tanB=2 then which one of the following is true ?

cos(A-B)=35 tan A tan B=2 sin A sin B=2cos A cos B cos (A-B)=35 cos A cos B+sin A sin B=35 cos A cos B+2cos A cos B=35 3cos A cos B=35 ⇒ cos A cos B=15 sin A sin B=25 cos(A+B)=cos A cos B-sin A sin B =15-25 =-15

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If $\cos (A - B) = \frac{3}{5}$ and $tanA tanB = 2$ then which one of the following is true ?

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$\cos (A - B) = \frac{- 1}{5}$

$\sin (A + B) = \frac{- 1}{5}$

$\sin (A + B) = \frac{1}{5}$

$\cos (A - B) = \frac{1}{5}$

answer is D.

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Detailed Solution

$\cos (A - B) = \frac{3}{5} \tan A \tan B = 2 \sin A \sin B = 2 \cos A \cos B \cos (A - B) = \frac{3}{5} \cos A \cos B + \sin A \sin B = \frac{3}{5} \cos A \cos B + 2 \cos A \cos B = \frac{3}{5} 3 \cos A \cos B = \frac{3}{5} \Rightarrow \cos A \cos B = \frac{1}{5} \sin A \sin B = \frac{2}{5} \cos (A + B) = \cos A \cos B - \sin A \sin B = \frac{1}{5} - \frac{2}{5} = \frac{- 1}{5}$