Question

If $\left( \frac { \sin \theta } { \sin \phi } \right) ^ { 2 } = \frac { \tan \theta } { \tan \phi } = 3$ then the value of $\theta$ and $\phi$ are

• A. $\theta = n \pi \pm \frac { \pi } { 3 } , \phi = n \pi \pm \frac { \pi } { 6 }$
• B. $\theta = n \pi - \frac { \pi } { 3 } , \phi = n \pi - \frac { \pi } { 6 }$
• C. $\theta = n \pi \pm \frac { \pi } { 2 } , \phi = n \pi + \frac { \pi } { 3 }$
• D. None of these

Answer 1

Answer: (A)

$\theta = n \pi \pm \frac { \pi } { 3 } , \phi = n \pi \pm \frac { \pi } { 6 }$

Answer 2

$\left( \frac { \sin \theta } { \sin \phi } \right) ^ { 2 } = \frac { \tan \theta } { \tan \phi }$ $\Rightarrow \sin \theta \cos \theta = \sin \phi \cos \phi$ $\Rightarrow \sin 2 \theta = \sin 2 \phi$

$2 \theta = \pi - 2 \phi$ $\Rightarrow \theta = \frac { \pi } { 2 } - \phi$

But, $\frac { \tan \theta } { \tan \phi } = 3$ $\Rightarrow \frac { \tan \theta } { \cot \theta } = 3$ $\Rightarrow \tan ^ { 2 } \theta = 3$ $\Rightarrow \theta = n \pi \pm \frac { \pi } { 3 }$ so

that $\phi = n \pi \pm \frac { \pi } { 6 }$

Trick: Check with the options for $n = 0 , n = 1$ .