A) \[\frac{18}{19}\]
B) \[\frac{3}{19}\]
C) \[\frac{1}{6}\]
D) None of these
Correct Answer: B
Solution :
Let\[AC=x=CB,\,AP=3AB=6x\]. Let \[2x=2n\pi \pm \left( \frac{\pi }{2}-2x \right)\] In \[\Delta ACP,\,\tan \alpha =\frac{x}{6x}=\frac{1}{6}\] In\[\Delta ABP\], \[\tan (\alpha +\beta )=\frac{2x}{6x}=\frac{1}{3}\] Now \[\tan \beta =\tan \{(\alpha +\beta )-\alpha \}=\frac{\tan (\alpha +\beta )-\tan \alpha }{1+\tan (\alpha +\beta )\tan \alpha }\] \[=\frac{\frac{1}{3}-\frac{1}{6}}{1+\frac{1}{3}.\frac{1}{6}}=\frac{1}{6}\times \frac{18}{19}=\frac{3}{19}\].You need to login to perform this action.
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