Key Concepts and Formulas

• Trigonometric Identities: $1 = \sin^2(x) + \cos^2(x)$, $\sin(2x) = 2\sin(x)\cos(x)$, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
• Inverse Trigonometric Functions: Understanding the principal value branch for $\tan^{-1}(\tan \theta)$ and when $\tan^{-1}(\tan \theta) = \theta$.
• Differentiation and its Applications: Finding the derivative of a function, calculating the slope of a tangent and a normal to a curve ($m_{normal} = -1/m_{tangent}$), and finding the equation of the normal using the point-slope form $y - y_1 = m(x - x_1)$.

Step-by-Step Solution

Step 1: Simplifying the Function $f(x)$

The given function is $f(x) = \tan^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)$, for $x \in \left(0, \frac{\pi}{2}\right)$. We aim to simplify this expression for easier differentiation.

Step 1.1: Express $1 \pm \sin x$ in terms of half-angle formulas.

We use the identities $1 = \sin^2(x/2) + \cos^2(x/2)$ and $\sin(x) = 2\sin(x/2)\cos(x/2)$. $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2\sin(x/2)\cos(x/2) = (\sin(x/2) + \cos(x/2))^2$ $1 - \sin x = \sin^2(x/2) + \cos^2(x/2) - 2\sin(x/2)\cos(x/2) = (\cos(x/2) - \sin(x/2))^2$

Why this step? This allows us to remove the square root in the next step.

Substituting these into $f(x)$: $f(x) = \tan^{-1}\left(\sqrt{\frac{(\sin(x/2) + \cos(x/2))^2}{(\cos(x/2) - \sin(x/2))^2}}\right)$

Step 1.2: Take the square root and analyze signs.

Since $x \in (0, \pi/2)$, we have $x/2 \in (0, \pi/4)$. In this interval, $\cos(x/2) > \sin(x/2) > 0$. Therefore, $\cos(x/2) - \sin(x/2) > 0$ and $\cos(x/2) + \sin(x/2) > 0$.

$f(x) = \tan^{-1}\left(\frac{\sin(x/2) + \cos(x/2)}{\cos(x/2) - \sin(x/2)}\right)$

Step 1.3: Convert the expression to terms of $\tan(x/2)$.

Divide both numerator and denominator by $\cos(x/2)$: $f(x) = \tan^{-1}\left(\frac{\frac{\sin(x/2)}{\cos(x/2)} + \frac{\cos(x/2)}{\cos(x/2)}}{\frac{\cos(x/2)}{\cos(x/2)} - \frac{\sin(x/2)}{\cos(x/2)}}\right) = \tan^{-1}\left(\frac{\tan(x/2) + 1}{1 - \tan(x/2)}\right)$

Step 1.4: Apply the tangent sum identity.

Using $\tan(\pi/4) = 1$, we rewrite the expression as: $f(x) = \tan^{-1}\left(\frac{\tan(\pi/4) + \tan(x/2)}{1 - \tan(\pi/4)\tan(x/2)}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right)$

Step 1.5: Simplify using $\tan^{-1}(\tan \theta) = \theta$.

For $x \in (0, \pi/2)$, we have $x/2 \in (0, \pi/4)$. Thus, $\frac{\pi}{4} + \frac{x}{2} \in (\frac{\pi}{4}, \frac{\pi}{2})$, which is within the principal value branch of $\tan^{-1}$, $(-\pi/2, \pi/2)$. Therefore, $f(x) = \frac{\pi}{4} + \frac{x}{2}$

Step 2: Finding the Derivative of $f(x)$

Now we differentiate the simplified function: $y = \frac{\pi}{4} + \frac{x}{2}$ $\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}\right) + \frac{d}{dx}\left(\frac{x}{2}\right) = 0 + \frac{1}{2} = \frac{1}{2}$

Step 3: Calculating the Slope of the Normal

The slope of the tangent is $m_t = \frac{dy}{dx} = \frac{1}{2}$. The slope of the normal is the negative reciprocal of the tangent's slope: $m_n = -\frac{1}{m_t} = -\frac{1}{\frac{1}{2}} = -2$

Step 4: Finding the Point on the Curve and Equation of the Normal

At $x = \frac{\pi}{6}$, the $y$-coordinate is: $y = f\left(\frac{\pi}{6}\right) = \frac{\pi}{4} + \frac{\pi/6}{2} = \frac{\pi}{4} + \frac{\pi}{12} = \frac{3\pi}{12} + \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$ So the point is $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$.

The equation of the normal at this point is: $y - \frac{\pi}{3} = -2\left(x - \frac{\pi}{6}\right)$ $y - \frac{\pi}{3} = -2x + \frac{\pi}{3}$ $y = -2x + \frac{2\pi}{3}$

Step 5: Checking Which Point Lies on the Normal

We need to find which of the given points satisfies the equation $y = -2x + \frac{2\pi}{3}$.

• (A) $\left(\frac{\pi}{6}, 0\right)$: $0 = -2\left(\frac{\pi}{6}\right) + \frac{2\pi}{3} = -\frac{\pi}{3} + \frac{2\pi}{3} = \frac{\pi}{3}$. This is false.
• (B) $\left(\frac{\pi}{4}, 0\right)$: $0 = -2\left(\frac{\pi}{4}\right) + \frac{2\pi}{3} = -\frac{\pi}{2} + \frac{2\pi}{3} = \frac{-\pi}{6}$. This is false.
• (C) $(0, 0)$: $0 = -2(0) + \frac{2\pi}{3} = \frac{2\pi}{3}$. This is false.
• (D) $\left(0, \frac{2\pi}{3}\right)$: $\frac{2\pi}{3} = -2(0) + \frac{2\pi}{3} = \frac{2\pi}{3}$. This is true.

Wait a minute! Something went wrong. Let's recheck option A: The normal equation is $y = -2x + \frac{2\pi}{3}$

Option (A) $\left( {{\pi \over 6},0} \right)$

$0 = -2 \cdot \frac{\pi}{6} + \frac{2\pi}{3} = -\frac{\pi}{3} + \frac{2\pi}{3} = \frac{\pi}{3}$ which is NOT TRUE.

Let's re-examine the question and the solution. We did everything correctly. The question MUST have an error. If we are given the fact that the answer is (A), then there must be an error in the normal equation.

Let's see what normal equation WE should have to make (A) the correct answer.

If (A) is on the line, then $0 = -2x + 2\pi/3$ should hold true for $x = \pi/6$.

Let's assume the equation is instead of $y = -2x + 2\pi/3$ is instead $y = mx + c$ Then, if $(\pi/6, 0)$ is on the line, we have $0 = m(\pi/6) + c$.

So we need to go back and re-derive the slope

$f(x) = \pi/4 + x/2$ $f'(x) = 1/2$ Normal slope = -2 Point on curve: $(\pi/6, f(\pi/6)) = (\pi/6, \pi/4 + \pi/12) = (\pi/6, 4\pi/12) = (\pi/6, \pi/3)$ Normal equation: $y - \pi/3 = -2(x - \pi/6)$

$y - \pi/3 = -2x + \pi/3$ $y = -2x + 2\pi/3$

Now if $(\pi/6, 0)$ is on the line then

$0 = -2(\pi/6) + 2\pi/3$ $0 = -\pi/3 + 2\pi/3 = \pi/3$, which is false.

Since everything is correct, the question is wrong.

Common Mistakes & Tips

• Always check the domain and range when simplifying inverse trigonometric functions.
• Remember that the slope of the normal is the negative reciprocal of the slope of the tangent.
• Double-check your calculations, especially when dealing with fractions and negative signs.

Summary

We simplified the given function using trigonometric identities and the properties of inverse trigonometric functions. We then found the derivative of the simplified function to determine the slope of the tangent and subsequently the slope of the normal. Finally, we derived the equation of the normal at the specified point and checked which of the given points lies on this normal. Since none of the options checked out, there must be an error in the question.

Final Answer The final answer is that based on our calculations, option (D) is actually correct, but the provided correct answer is A, which is incorrect. There seems to be an error in the question's specified correct answer.