$I f \cos A = - 3 / 5, \sin B = 7 / 25 a n d 90^{o} < A < 180^{o}, 0^{o} < B < 90^{o}, t h e n \tan (A + B) =$

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3 /4

-3 /4

3/5

-3 / 5

answer is B.

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Detailed Solution

$\begin{array}{l} \cos A = - 3 / 5, A \in I I \Rightarrow \tan A = - 4 / 3 \\ \sin B = 7 / 25, B \in I \Rightarrow \tan B = 7 / 24. ∴ \tan (A + B) = - 3 / 4 \end{array}$

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