Question

If \[\cos \theta = \dfrac{2}{3}\] and $\theta $ is in the 4th quadrant, then $\dfrac{{\sin \theta + \tan \theta }}{{\sin \theta - \tan \theta }}$ is equal to
(A) $\dfrac{{ - 1}}{5}$
(B) $\dfrac{1}{5}$
(C) $ - 5$
(D) $5$

Answer 1

Hint:
First of all find the value of $\sin \theta $ by using the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and then find $\tan \theta $ by using, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ to evaluate $\dfrac{{\sin \theta + \tan \theta }}{{\sin \theta - \tan \theta }}$.

Complete step by step solution:
Given, \[\cos \theta = \dfrac{2}{3}\] and $\theta $ is in the 4^th quadrant.
To evaluate $\dfrac{{\sin \theta + \tan \theta }}{{\sin \theta - \tan \theta }}$; we will need to find the value of $\sin \theta $ and $\tan \theta $.
Now, find the value of $\sin \theta $by using the identity:
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$ \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta $
$ \Rightarrow \sin \theta = \pm \sqrt {1 - {{\cos }^2}\theta } $
We have given that $\theta $ is in the 4^th quadrant. Therefore, $\sin \theta $will be negative.
$\sin \theta = - \sqrt {1 - {{\cos }^2}\theta } $
$ \Rightarrow \sin \theta = - \sqrt {1 - {{\left( {\dfrac{2}{3}} \right)}^2}} $
$ \Rightarrow \sin \theta = - \sqrt {1 - \dfrac{4}{9}} $
$ \Rightarrow \sin \theta = - \sqrt {\dfrac{{9 - 4}}{9}} $
$ \Rightarrow \sin \theta = \dfrac{{ - \sqrt 5 }}{3}$
Now, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$ \Rightarrow \tan \theta = \dfrac{{\dfrac{{ - \sqrt 5 }}{3}}}{{\dfrac{2}{3}}}$
$ \Rightarrow \tan \theta = \dfrac{{ - \sqrt 5 }}{2}$
$\therefore \dfrac{{\sin \theta + \tan \theta }}{{\sin \theta - \tan \theta }} = \dfrac{{\dfrac{{ - \sqrt 5 }}{3} + \left( {\dfrac{{ - \sqrt 5 }}{2}} \right)}}{{\dfrac{{ - \sqrt 5 }}{3} - \left( {\dfrac{{ - \sqrt 5 }}{2}} \right)}}$
$ = \dfrac{{\dfrac{{ - \sqrt 5 }}{3} - \dfrac{{\sqrt 5 }}{2}}}{{\dfrac{{ - \sqrt 5 }}{3} + \dfrac{{\sqrt 5 }}{2}}}$
\[ = \dfrac{{\dfrac{{ - 2\sqrt 5 - 3\sqrt 5 }}{6}}}{{\dfrac{{ - 2\sqrt 5 + 3\sqrt 5 }}{6}}}\]
\[ = \dfrac{{ - 5\sqrt 5 }}{{\sqrt 5 }}\]
$ = - 5$
$\therefore \dfrac{{\sin \theta + \tan \theta }}{{\sin \theta - \tan \theta }} = - 5$

Hence, option (C) is the correct answer.

Note:
By ASTC formula, only $\cos \theta $ and $\sec \theta $ are positive in 4^th quadrant while other trigonometry ratios $\left( {\sin \theta ,\tan \theta ,\cot \theta ,\cos ec\theta } \right)$are negative in 4^th quadrant.