A) \[\frac{23}{16}\]
B) \[\frac{16}{63}\]
C) \[\frac{1}{63}\]
D) \[\frac{13}{63}\]
Correct Answer: B
Solution :
\[\sin A=\frac{3}{5},\] \[\tan A=\frac{\sin A}{\sqrt{1-{{\sin }^{2}}A}}=\frac{35}{\sqrt{1-{{(3/5)}^{2}}}}\] \[=\frac{\frac{3}{5}}{\sqrt{1-\frac{9}{25}}}=\frac{3/5}{4/5}=\frac{3}{4}\] \[\cos B=\frac{12}{13},\] \[\tan B=\frac{1-{{\cos }^{2}}B}{\cos B}\] \[=\frac{\sqrt{1-\frac{144}{169}}}{\frac{12}{13}}=\frac{5/13}{12/13}=\frac{5}{12}\] \[\therefore \] \[\frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{\frac{3}{4}-\frac{5}{12}}{1+\frac{3}{4}-\frac{5}{12}}\] \[=\frac{\frac{9-5}{12}}{\frac{48+15}{4\times 12}}=\frac{\frac{1}{3}}{\frac{63}{48}}\] \[=\frac{1}{3}\times \frac{48}{63}=\frac{16}{63}\]
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