Question

If \[\frac{3\pi}{4} < \alpha < \pi, \text{ then }\sqrt{2\cot \alpha + \frac{1}{\sin^2 \alpha}}\] is equal to

विकल्प

• 1 − cot α

• 1 + cot α

• −1 + cot α

• −1 −cot α

Answer 1

−1 −cot α

We have: 

\[ \sqrt{2\cot\alpha + \frac{1}{\sin^2 \alpha}} \]

\[ = \sqrt{\frac{2\cos\alpha}{\sin\alpha} + \frac{1}{\sin^2 \alpha}}\]

\[ = \sqrt{\frac{2\sin \alpha\cos \alpha + 1}{\sin^2 \alpha}}\]

\[ = \sqrt{\frac{2\sin \alpha\cos\alpha + \sin^2 \alpha + \cos^2 \alpha}{\sin^2 \alpha}}\]

\[ = \sqrt{\frac{\left( \sin\alpha + \cos\alpha \right)^2}{\sin^2 \alpha}}\]

\[ = \sqrt{\left( 1 + \cot \alpha \right)^2}\]

\[ = \left| 1 + \cot \alpha \right|\]

\[ = - \left( 1 + \cot \alpha \right) \left[ \text{ When } \frac{3\pi}{4} < \alpha < \pi, \cot \alpha < - 1 \Rightarrow \cot \alpha + 1 < 0 \right]\]

\[ = - 1-\cot \alpha\]