A) \[\frac{14}{15}\]
B) \[\frac{25}{34}\]
C) \[\frac{13}{25}\]
D) \[\frac{11}{46}\]
Correct Answer: B
Solution :
(b): \[\tan A+\operatorname{tanB}=\frac{5}{3}\]and\[\tan A\operatorname{tanB}=2\] \[\tan (A+B)=W=\frac{-5}{3}\] \[\therefore {{\sin }^{2}}(A+B)=\frac{1}{\cos e{{c}^{2}}A}=\frac{1}{1+{{\cot }^{2}}B}\] \[=\frac{1}{1+\frac{9}{25}}=\frac{25}{34}\]You need to login to perform this action.
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