A) 4
B) \[\frac{4}{3}\]
C) \[-\frac{1}{4}\]
D) -4
Correct Answer: A
Solution :
| [a] |
| Converting \[tan\,\alpha \] and \[cot\,\alpha \] in \[sin\,\alpha \]and \[cos\,\,\alpha \]: |
| \[y=\sqrt{2\cot \alpha +\cos e{{c}^{2}}\alpha }=\sqrt{2\cot \alpha +1+{{\cot }^{2}}\alpha }\] |
| \[=\left| \cot \alpha +1 \right|\] |
| As \[\alpha \in \left( \frac{3\pi }{4},\pi \right)\] |
| \[y=-1-\cot \alpha \] |
| \[\frac{dy}{d\alpha }=\cos e{{c}^{2}}\alpha \] |
| At \[\alpha =\frac{5\pi }{6},\frac{dy}{d\alpha }=4\] |
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