Question

If $\sin \theta +\cos \theta =\frac{1}{5}$ and $0\le \theta <\pi$, then $\tan \theta$ is

• A. $-\frac{4}{3}$
• B. $-\frac{2}{4}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $-\frac{4}{3}$

$\sin \theta +\cos \theta =\frac{1}{5}$

Squaring both side, we get

$\Rightarrow 1+\sin 2\theta =\frac{1}{25}$

$\Rightarrow \sin 2\theta =\frac{-24}{25}\Rightarrow \frac{2\tan \theta }{1+{\tan }^2\theta }=\frac{-24}{25}$

$\Rightarrow 12{\tan }^2\theta +25\tan \theta +12=0$

$\Rightarrow (3\tan \theta +4)(4\tan \theta +3)=0$

$\Rightarrow \tan \theta =\frac{-3}{4}$ or $\frac{-4}{3}$