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Question: If \(\cos(A - B) = \frac{3}{5}\) and \(\tan A\tan B = 2,\) then...

If cos⁡(A−B)=35\cos(A - B) = \frac{3}{5} and tan⁡Atan⁡B=2,\tan A\tan B = 2, then

A

cos⁡Acos⁡B=15\cos A\cos B = \frac{1}{5}

B

sin⁡Asin⁡B=−25\sin A\sin B = - \frac{2}{5}

C

cos⁡Acos⁡B=−15\cos A\cos B = - \frac{1}{5}

D

sin⁡Asin⁡B=−15\sin A\sin B = - \frac{1}{5}

Answer

cos⁡Acos⁡B=15\cos A\cos B = \frac{1}{5}

Explanation

Solution

cos⁡(A−B)=35\cos(A - B) = \frac{3}{5}

∴ 5cos⁡Acos⁡B+5sin⁡Asin⁡B=35\cos A\cos B + 5\sin A\sin B = 3 …..(i)

From 2nd relation,

sin⁡Asin⁡B=2cos⁡Acos⁡B\sin A\sin B = 2\cos A\cos B .....(ii)

∴\therefore cos⁡Acos⁡B=15\cos A\cos B = \frac{1}{5} and 5(12+1)sin⁡Asin⁡B=35\left( \frac{1}{2} + 1 \right)\sin A\sin B = 3.