Question

$\sec \theta -\tan \theta =3\Rightarrow \theta$ lies in the quadrant

• A. $I$
• B. $II$
• C. $III$
• D. $IV$

Answer 1

The correct option is D $IV$

$\sec \theta -\tan \theta =3$
$weknow$
${\sec }^2\theta -{\tan }^2\theta =1$
$\Rightarrow (\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=1$
$\Rightarrow \sec \theta +\tan \theta =\frac{1}{\sec \theta -\tan \theta }$
$\Rightarrow \sec \theta +\tan \theta =\frac{1}{3}......\left(i\right)$
$\sec \theta -\tan \theta =3.......\left(ii\right)\left(given\right)$
$adding\left(i\right)and\left(ii\right)$
$2\sec \theta =\frac{10}{3}$
$\sec \theta =\frac{5}{3}$
$\sec \theta -\tan \theta =3$
$\frac{5}{3}-\tan \theta =3$
$\tan \theta =\frac{5}{3}-3$
$=-\frac{4}{3}$
$\sin ce\sec \theta is+ve,\tan \theta is-ve$
$Hence\theta liesinIVquadrant$