Question

If \( y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} ,\alpha \in \left( {{{3\pi } \over 4},\pi } \right) \)

\( {{dy} \over {d\alpha }}\,\,at\,\alpha = {{5\pi } \over 6}is \) :

• A. 4
• B. -4
• C. \( {4 \over 3} \)
• D. -\( {1 \over 4} \)

Hint:

Simplify the expression under the root to a perfect square using \(\tan\alpha + \cot\alpha = \frac{1}{\sin\alpha\cos\alpha}\), then watch the sign of 1 + cot \(\alpha\) on the given interval.

Answer 1

The Correct Option is A

\(\frac{\tan\alpha + \cot\alpha}{1 + \tan^2\alpha} = \frac{1/(\sin\alpha\cos\alpha)}{\sec^2\alpha} = \frac{\cos^2\alpha}{\sin\alpha\cos\alpha} = \cot\alpha\).
Under the root: \(2\cot\alpha + \frac{1}{\sin^2\alpha} = 2\cot\alpha + 1 + \cot^2\alpha = (1 + \cot\alpha)^2\).
So \(y = |1 + \cot\alpha|\). For \(\alpha \in (3\pi/4, \pi)\), \(\cot\alpha \lt -1\), so \(1 + \cot\alpha \lt 0\) and \(y = -(1 + \cot\alpha)\).
\(\frac{dy}{d\alpha} = \csc^2\alpha\). At \(\alpha = 5\pi/6\), \(\sin\alpha = \tfrac12\), so \(\frac{dy}{d\alpha} = 4\).
Answer: (A) 4