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)

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