A) \[\frac{56}{33}\]
B) \[\frac{56}{23}\]
C) \[\frac{43}{33}\]
D) \[\frac{34}{33}\]
Correct Answer: A
Solution :
| \[=\text{ }12+4=16\text{ cm}\] \[=\frac{3}{5}\] |
| \[\cos \,\,(\alpha -\beta )=\sqrt{1-{{\sin }^{2}}\,\,(\alpha -\beta )}=\sqrt{1-\frac{25}{169}}\]\[=\sqrt{\frac{144}{169}}=\frac{12}{13}\] |
| \[\therefore \]\[\tan \,\,(\alpha +\beta )=\frac{\sin \,\,(\alpha +\beta )}{\cos \,\,(\alpha +\beta )}=\left( \frac{3}{5}\times \frac{5}{4} \right)=\frac{3}{4}\] |
| \[\tan \,\,(\alpha +\beta )=\frac{\sin \,\,(\alpha -\beta )}{\cos \,\,(\alpha -\beta )}=\frac{5}{13}\times \frac{13}{12}=\frac{5}{12}\] |
| \[\therefore \]\[\tan \,\,(2\alpha )=\tan [(\alpha +\beta )+(\alpha -\beta )]\] |
| \[=\frac{\tan \,\,(\alpha +\beta )+\tan \,\,(\alpha -\beta )}{1-\tan \,\,(\alpha +\beta )\cdot \tan \,\,(\alpha -\beta )}=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\times \frac{5}{12}}=\frac{56}{33}\] |
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