A) \[\frac{1-\sqrt{7}}{4}\]
B) \[\frac{4-\sqrt{7}}{3}\]
C) \[-\left( \frac{4+\sqrt{7}}{3} \right)\]
D) \[\frac{1+\sqrt{7}}{4}\]
Correct Answer: C
Solution :
[c]| Given, \[\sin x+\cos x=\frac{1}{2},x\in (0,\pi )\] |
| On squaring both sides, we get \[1+\sin 2x=\frac{1}{4}\] |
| \[\Rightarrow \]\[\sin 2x=\frac{1}{4}-1=-\frac{3}{4}<0\] |
| \[\therefore \]\[x\in \left( \frac{\pi }{2},\pi \right)\] |
| \[\cos 2x=-\,\sqrt{1-{{\sin }^{2}}2x}\] |
| \[=-\sqrt{1-\frac{9}{16}}=-\frac{\sqrt{7}}{4}\] |
| \[\tan x=\frac{1-\cos 2x}{\sin 2x}\]\[=\frac{1+\frac{\sqrt{7}}{4}}{\frac{-\,3}{4}}=-\left( \frac{4+\sqrt{7}}{3} \right)\] |
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