Question

If \(\cos\theta\cosec\theta=-1\) and \(\theta\) lies in the second quadrant, then \(\cos\theta=\)

• A. \(-\frac{\sqrt{3}}{2}\)
• B. \(\frac{\sqrt{2}}{2}\)
• C. \(-\frac{\sqrt{2}}{2}\)
• D. \(-\sqrt{2}\)

Hint:

In the second quadrant, sine is positive and cosine is negative. If \(\tan\theta=-1\), then \(\theta=135^\circ\), so \(\cos\theta=-\frac{\sqrt{2}}{2}\).

Answer 1

The Correct Option is C

We are given \[ \cos\theta\cosec\theta=-1. \] We know that \[ \cosec\theta=\frac{1}{\sin\theta}. \] Therefore, \[ \cos\theta\cosec\theta = \cos\theta\cdot \frac{1}{\sin\theta}. \] \[ \cos\theta\cosec\theta=\frac{\cos\theta}{\sin\theta}. \] But, \[ \frac{\cos\theta}{\sin\theta}=\cot\theta. \] So, \[ \cot\theta=-1. \] Hence, \[ \tan\theta=-1. \] Now \(\theta\) lies in the second quadrant. In the second quadrant: \[ \sin\theta>0,\quad \cos\theta<0,\quad \tan\theta<0. \] The angle for which \(\tan\theta=-1\) in the second quadrant is \[ \theta=135^\circ=\frac{3\pi}{4}. \] Therefore, \[ \cos\theta=\cos 135^\circ. \] \[ \cos 135^\circ=-\frac{1}{\sqrt{2}}. \] \[ \cos\theta=-\frac{\sqrt{2}}{2}. \]