Question

If $\sec \theta+\tan \theta=1 / 5$, then the value of $\sin \theta$ is

• A. $\frac{12}{13}$
• B. $-\frac{12}{13}$
• C. $\pm \frac{12}{13}$
• D. $\frac{5}{12}$

Answer 1

Answer: (B)

$\sec \theta+\tan \theta=\frac{1}{5}$
$\Rightarrow \frac{1+\sin \theta}{\cos \theta}=\frac{1}{5}$
$ \Rightarrow(1+\sin \theta)=\frac{1}{5} \cos \theta $
$\text { squaring, }(1+\sin \theta)^2=\frac{1}{25} \cos ^2 \theta$
$\Rightarrow 25+25 \sin ^2 \theta+50 \sin \theta=1-\sin ^2 \theta $
$\Rightarrow 13 \sin ^2 \theta+25 \sin \theta+12=0 $
$\Rightarrow(13 \sin \theta+12)(\sin \theta+1)=0$
$\Rightarrow \sin \theta=-1 \text { or } \sin \theta=\frac{-12}{13} $
$\Rightarrow \theta=(4 n-1) \frac{\pi}{2}, n \in I $
$\therefore \sin \theta=\frac{-12}{13}$
which is not possible
$\because \cos \theta=0 $