Question

The most general value of $\theta$which will satisfy both the

equations $\sin \theta = \frac { - 1 } { 2 }$and $\tan \theta = \frac { 1 } { \sqrt { 3 } }$ is

• A. $n \pi + ( - 1 ) ^ { n } \frac { \pi } { 6 }$
• B. $n \pi + \frac { \pi } { 6 }$
• C. $2 n \pi \pm \frac { \pi } { 6 }$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

$\sin \theta = \frac { - 1 } { 2 } = \sin \left( \frac { - \pi } { 6 } \right) = \sin \left( \pi + \frac { \pi } { 6 } \right)$ $\tan \theta = \left( \frac { 1 } { \sqrt { 3 } } \right) = \tan \left( \frac { \pi } { 6 } \right) = \tan \left( \pi + \frac { \pi } { 6 } \right)$ $\Rightarrow \theta = \left( \pi + \frac { \pi } { 6 } \right)$

Hence, general value of $\theta$ is $2 n \pi + \frac { 7 \pi } { 6 }$