Question

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

• A. $-4/3$
• B. $-3/4$
• C. $3/4$
• D. $4/3$

Answer 1

Answer: (A)

$\cos \theta +\sin \theta =\frac{1}{5}$
$\Rightarrow 1+\tan \theta =\frac{1}{5}\sec \theta$ (dividing by $\cos \theta$ )
$\Rightarrow 25+50\tan \theta +25{\tan }^2\theta ={\sec }^2\theta$
or $12{\tan }^2\theta +25\tan \theta +12=0$
or $12{\tan }^2\theta +16\tan \theta +9\tan \theta +12=0$
or $(4\tan \theta +3)(3\tan \theta +4)=0$
$\Rightarrow \tan \theta =-\frac{3}{4}$
or $\tan \theta =-\frac{4}{3}$
$\tan \theta =-\frac{3}{4}$ is rejected as then $\cos \theta =-\frac{4}{5}$ and $\sin \theta =\frac{3}{5}$
for which $\cos \theta +\sin \theta =-\frac{1}{5}$
Hence, $\tan \theta =-\frac{4}{3}$