If \(y(x) = \cot^{-1}\left(\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right), x \in \left(\frac{\pi}{2}, \pi\right)\), then \(\frac{dy}{dx}\) at \(x = \frac{5\pi}{6}\) is :
Show Hint
The key to simplifying expressions like \(\sqrt{1 \pm \sin x}\) is using the perfect square identities. Always be careful about the absolute value, \( \sqrt{A^2} = |A| \), and check the sign of the expression inside the absolute value based on the given interval for x.
Step 1: Simplify the expression inside the cot\(^{-1}\).
We use the identities:
\[ 1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2 \]
\[ 1 - \sin x = \cos^2(x/2) + \sin^2(x/2) - 2\sin(x/2)\cos(x/2) = (\cos(x/2) - \sin(x/2))^2 \]
Therefore,
\[ \sqrt{1 + \sin x} = |\cos(x/2) + \sin(x/2)| \]
\[ \sqrt{1 - \sin x} = |\cos(x/2) - \sin(x/2)| \]
Step 2: Analyze the signs in the given interval.
The interval for x is \((\pi/2, \pi)\).
This means the interval for \(x/2\) is \((\pi/4, \pi/2)\).
In the interval \((\pi/4, \pi/2)\):
- Both \(\cos(x/2)\) and \(\sin(x/2)\) are positive. So, \(\cos(x/2) + \sin(x/2)\) is positive.
- \(\sin(x/2)>\cos(x/2)\). So, \(\cos(x/2) - \sin(x/2)\) is negative.
Therefore,
\[ \sqrt{1 + \sin x} = \cos(x/2) + \sin(x/2) \]
\[ \sqrt{1 - \sin x} = -(\cos(x/2) - \sin(x/2)) = \sin(x/2) - \cos(x/2) \]
Step 3: Substitute the simplified terms back.
Let E be the expression inside \(\cot^{-1}\).
\[ E = \frac{(\cos(x/2) + \sin(x/2)) + (\sin(x/2) - \cos(x/2))}{(\cos(x/2) + \sin(x/2)) - (\sin(x/2) - \cos(x/2))} \]
\[ E = \frac{2\sin(x/2)}{2\cos(x/2)} = \tan(x/2) \]
So, \(y(x) = \cot^{-1}(\tan(x/2))\). Step 4: Convert tan to cot.
We use the identity \(\tan\theta = \cot(\pi/2 - \theta)\).
\[ y(x) = \cot^{-1}(\cot(\pi/2 - x/2)) \]
For \(x \in (\pi/2, \pi)\), we have \(x/2 \in (\pi/4, \pi/2)\).
Then \(\pi/2 - x/2 \in (0, \pi/4)\).
Since \((\pi/2 - x/2)\) lies in the principal value range of \(\cot^{-1}\) (which is \((0, \pi)\)), we can write:
\[ y(x) = \pi/2 - x/2 \]
Step 5: Differentiate and evaluate.
\[ \frac{dy}{dx} = \frac{d}{dx}(\pi/2 - x/2) = -1/2 \]
The derivative is a constant, so its value at \(x = 5\pi/6\) is also -1/2.
