Question

Value of $ \theta (0 < \theta < 360^\circ ) $ which satisfy the equation $ \csc \theta + 2 = 0 $
(A) $ 210^\circ ,100^\circ $
(B) $ 240^\circ ,300^\circ $
(C) $ 210^\circ ,240^\circ $
(D) $ 210^\circ ,330^\circ $

Answer 1

Hint: Simplify the equation $ \csc \theta + 2 = 0 $ to get $ \sin \theta = - \dfrac{1}{2} $ . Use the fact that $ \sin 30^\circ = \dfrac{1}{2} $ and $ \sin ( - x) = - \sin x $ . Compute \[\sin (180 + 30)^\circ \] and \[\sin (360 - 30)^\circ \] to get the answer.

Complete step-by-step answer:
We are given the equation $ \csc \theta + 2 = 0 $ .
We need to find a value of $ \theta $ such that $ \csc \theta + 2 = 0 $ and $ 0 < \theta < 360^\circ $ .
Consider the equation $ \csc \theta + 2 = 0 $ .
Then $ \csc \theta = - 2.......(1) $
We know that $ \sin \theta = \dfrac{1}{{\csc \theta }} $ .
Taking reciprocal on both sides, we get
  $
  \dfrac{1}{{\csc \theta }} = \dfrac{1}{{ - 2}} \\
   \Rightarrow \sin \theta = - \dfrac{1}{2} \\
   $
Here $ \sin \theta $ is negative, therefore, $ \theta $ will be in the third or fourth quadrant.

We know that $ \sin 30^\circ = \dfrac{1}{2} $ , $ \sin ( - x) = - \sin x $ for any $ 0 < x < 360^\circ $ .
Therefore, \[\sin ( - 30)^\circ = - \dfrac{1}{2}\]
We know that sine is a periodic function with its period being $ 360^\circ $ or $ 2\pi $ .
Therefore, \[\sin ( - 30 + n360)^\circ = - \dfrac{1}{2}\] for any integer n but we have the condition that $ 0 < \theta < 360^\circ $
Therefore n = 1, and \[\sin (360 - 30)^\circ = - \dfrac{1}{2} = \sin 330^\circ \].
Also, $ \sin (180 + x) = - \sin x $ for any $ 0 < x < 360^\circ $ .
Therefore, \[\sin (180 + 30)^\circ = - \sin 30^\circ = - \dfrac{1}{2} = \sin 210^\circ \]
Hence the value of $ \theta $ is $ 210^\circ ,330^\circ $ .
So, the correct answer is “Option D”.


Note: Identifying the quadrants where the value of $ \theta $ lies is one of the crucial steps to solving such questions. The value of $ \sin \theta $ is positive in the first and second quadrants and negative in the third and fourth quadrants. The value of $ \cos \theta $ is positive in the first and the fourth quadrants and negative in the remaining ones.