If 270∘<A<360∘,90∘<B<180∘ and cosA=513, tanB=−158 then sin(A+B)=

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If $270^{\circ} < A < 360^{\circ}, 90^{\circ} < B < 180^{\circ}$ and $cosA = \frac{5}{13}, tanB = - \frac{15}{8}$ then $\sin (A + B) =$

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$\frac{140}{221}$

$\frac{140}{171}$

$\frac{171}{221}$

$\frac{221}{171}$

answer is B.

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Detailed Solution

Given $270^{\circ} < A < 360^{\circ}$
$\begin{array}{l} 90^{\circ} < B < 180^{\circ} \\ A \in IV - quad ant \\ B \in II - quadrant \\ \cos A = \frac{5}{13} \Rightarrow \sin A = - \frac{12}{13} a n d \cos B = \frac{- 8}{17}, \sin B = \frac{15}{17} \\ G.E = \sin Acos B + \cos Asin B = (\frac{- 12}{13}) (\frac{- 8}{17}) + (\frac{5}{13}) (\frac{15}{17}) \\ = \frac{96 + 75}{221} = \frac{171}{221} \end{array}$