A) \[\frac{(4-\sqrt{7})}{3}\]
B) \[-\frac{(4+\sqrt{7})}{3}\]
C) \[\frac{(1+\sqrt{7})}{4}\]
D) \[\frac{(1-\sqrt{7})}{4}\]
Correct Answer: B
Solution :
Since, \[\cos x+\sin x=\frac{1}{2}\] ...(i) \[\Rightarrow \] \[\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}+\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}=\frac{1}{2}\] Put \[\tan \frac{x}{2}=t\] \[\therefore \] \[\frac{1-{{t}^{2}}}{1+{{t}^{2}}}+\frac{2t}{1+{{t}^{2}}}=\frac{1}{2}\] \[\Rightarrow \] \[3{{t}^{2}}-4t-1=0\] \[\therefore \] \[t=\frac{2\pm \sqrt{7}}{3}\] As \[0<x<\pi \Rightarrow 0<\frac{x}{2}<\frac{x}{2}\] \[\therefore \]\[\tan \frac{x}{2}\]is positive. \[\therefore \]We take, \[t=\tan \frac{x}{2}=\left( \frac{2+\sqrt{7}}{3} \right)\] Since, \[\tan x=\frac{2\tan \frac{x}{2}}{1-{{\tan }^{2}}\frac{x}{2}}=\frac{2t}{1-{{t}^{2}}}\] \[=\frac{2\left( \frac{2+\sqrt{7}}{3} \right)}{1-{{\left( \frac{2+\sqrt{7}}{3} \right)}^{2}}}=\frac{-3(2+\sqrt{7})}{(1+2\sqrt{7})}\times \frac{(1-2\sqrt{7})}{(1-2\sqrt{7})}\] \[\therefore \]\[\tan x=-\left( \frac{4+\sqrt{7}}{3} \right)\]
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