Trigonometric Identities Question 231
Question: If $ 0<x<\pi $ it and $ \cos x+\sin x=\frac{1}{2}, $ then tan x is
Options:
A) $ \frac{(1-\sqrt{7})}{4} $
B) $ \frac{(4-\sqrt{7})}{3} $
C) $ -\frac{(4+\sqrt{7})}{3} $
D) $ \frac{(1+\sqrt{7})}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
Given $ \cos x+\sin x=\frac{1}{2}\Rightarrow 1+\sin 2x=\frac{1}{4} $
$ \Rightarrow \sin 2x=-\frac{3}{4}, $ so x is obtuse and $ \frac{2\tan x}{1+{{\tan }^{2}}x}=\frac{3}{4} $
$ \Rightarrow 3{{\tan }^{2}}x+8\tan x+3=0 $
$ \therefore \tan x=\frac{-8\pm \sqrt{64-36}}{6}=-\frac{-4\pm \sqrt{7}}{3} $ as $ \tan x<0 $
$ \therefore \tan x=\frac{-4-\sqrt{7}}{3} $
BETA
