Question

If $0<x<\pi$ and $\cos x+\sin x=\frac{1}{2}$ then prove that $\tan x=-\left(\frac{4+\sqrt{7}}{3}\right)$.

Answer 1

$\begin{array}{c}cosx+sinx=\frac{1}{2}....\left(1\right)\\ prove\tan x=-\left(\frac{4+\sqrt{7}}{3}\right)\\ Squaringbothsideofeq\left(1\right)\\ \Rightarrow {\left(\cos x+\sin x\right)}^2={\left(\frac{1}{2}\right)}^2\left[\because 2\sin \theta \cdot \cos \theta =\sin 2\theta and{\sin }^2\theta +{\cos }^2\theta =1\right]\\ \Rightarrow \left({\sin }^{2}x+{\cos }^{2}x\right)+2\sin x.\cos x=\frac{1}{4}\\ \Rightarrow 1+\sin 2x=\frac{1}{4}\\ \Rightarrow \sin 2x=\frac{1}{4}-1=-\frac{3}{4}\\ \Rightarrow \frac{2\tan x}{1+{\tan }^{2}x}=-\frac{3}{4}\\ \Rightarrow -\frac{3}{2}\tan x=\left(1+{\tan }^{2}x\right)\\ \Rightarrow -3\tan x=2+2{\tan }^{2}x\\ \Rightarrow 2{\tan }^{2}x+3\tan x+2=0\\ D=9-4\times 2\times 2=-7\left[factorbydiscriminentmethod\right]\\ then\\ Roots=\frac{-3\pm \sqrt{-7}}{2\times 2}\\ \therefore \tan x=\frac{-\left(3+\sqrt{-7}\right)}{4}or\frac{\left(\sqrt{-7}-3\right)}{4}\end{array}$