Step 1: Application of Trigonometric Half-Angle Identities

To simplify the expression inside the inverse tangent function, we utilize the power-reduction identities derived from the cosine double-angle formulas:

• $1 - \cos x = 2 \sin^2 \left( \frac{x}{2} \right)$ [Per the identity $\cos 2\theta = 1 - 2\sin^2 \theta$]
• $1 + \cos x = 2 \cos^2 \left( \frac{x}{2} \right)$ [Per the identity $\cos 2\theta = 2\cos^2 \theta - 1$]

Step 2: Algebraic Substitution and Simplification

Substitute the identities into the given expression:

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

Canceling the constant factor of 2:

$= \sqrt{\frac{\sin^2 \left( \frac{x}{2} \right)}{\cos^2 \left( \frac{x}{2} \right)}} = \sqrt{\tan^2 \left( \frac{x}{2} \right)}$

$= \left| \tan \left( \frac{x}{2} \right) \right|$

Step 3: Evaluating the Domain Constraint

Given the constraint $0 < x < \pi$, we divide the inequality by 2 to determine the range of the argument:

$0 < \frac{x}{2} < \frac{\pi}{2}$

In the interval $(0, \frac{\pi}{2})$, the tangent function is positive. Therefore, $\left| \tan \left( \frac{x}{2} \right) \right| = \tan \left( \frac{x}{2} \right)$.

Step 4: Final Inverse Trigonometric Operation

Substitute the simplified internal term back into the inverse function:

$\tan^{-1} \left( \tan \left( \frac{x}{2} \right) \right)$

Since $\frac{x}{2} \in (0, \frac{\pi}{2})$, which lies within the principal value branch of $\tan^{-1} x$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$, we apply the identity $\tan^{-1}(\tan \theta) = \theta$:

$\tan^{-1} \left( \tan \left( \frac{x}{2} \right) \right) = \frac{x}{2}$

Final Solution: \frac{x}{2}

Step 1: Identifying the Trigonometric Substitution

To simplify the expression $\tan^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right)$, we observe the form $\sqrt{a^2 - x^2}$, which suggests a substitution based on the Pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$, specifically $1 - \sin^2 \theta = \cos^2 \theta$.

Let $x = a \sin \theta$. Given the constraint $|x| < a$, we have $\left| \frac{x}{a} \right| < 1$, which implies $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$.

[By the substitution method for inverse trigonometric integrals and simplifications]

Step 2: Substitution and Algebraic Simplification

Substitute $x = a \sin \theta$ into the given expression:

$\tan^{-1} \left( \frac{a \sin \theta}{\sqrt{a^2 - (a \sin \theta)^2}} \right)$

Factor out $a^2$ within the square root:

$= \tan^{-1} \left( \frac{a \sin \theta}{\sqrt{a^2(1 - \sin^2 \theta)}} \right)$

Apply the trigonometric identity $1 - \sin^2 \theta = \cos^2 \theta$:

$= \tan^{-1} \left( \frac{a \sin \theta}{\sqrt{a^2 \cos^2 \theta}} \right)$ [Per the Identity: $\sin^2 \theta + \cos^2 \theta = 1$]

Step 3: Evaluating the Square Root and Tangent Conversion

Since $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $\cos \theta > 0$. Therefore, $\sqrt{a^2 \cos^2 \theta} = a \cos \theta$ (assuming $a > 0$):

$= \tan^{-1} \left( \frac{a \sin \theta}{a \cos \theta} \right)$ $= \tan^{-1} (\tan \theta)$ [By the quotient identity: $\tan \theta = \frac{\sin \theta}{\cos \theta}$]

Step 4: Final Domain Analysis

Since $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, the inverse function property $\tan^{-1}(\tan \theta) = \theta$ holds true.

From the initial substitution $x = a \sin \theta$, we solve for $\theta$:

$\frac{x}{a} = \sin \theta \implies \theta = \sin^{-1} \left( \frac{x}{a} \right)$

Final Solution: \sin^{-1} \left( \frac{x}{a} \right)

Step 1: Algebraic Manipulation of the Trigonometric Expression

Given the expression: $f(x) = \tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right)$.

To simplify the argument of the inverse tangent function, divide both the numerator and the denominator by $\cos x$ [Justification: This transformation is valid as $\cos x \neq 0$ within the domain $\frac{-\pi}{4} < x < \frac{3\pi}{4}$].

$$\frac{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}} = \frac{1 - \tan x}{1 + \tan x}$$

Step 2: Applying Trigonometric Identities

Recall the trigonometric identity for the tangent of a difference: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

We can rewrite the expression $1 - \tan x$ by substituting $1$ with $\tan \frac{\pi}{4}$:

$$\frac{\tan \frac{\pi}{4} - \tan x}{1 + (\tan \frac{\pi}{4})(\tan x)}$$

[Per the Addition/Subtraction Formula for Tangent, this simplifies to $\tan(\frac{\pi}{4} - x)$].

Step 3: Simplification via Inverse Trigonometric Properties

Now, the function becomes: $f(x) = \tan^{-1} \left( \tan \left( \frac{\pi}{4} - x \right) \right)$.

We must check if the argument $(\frac{\pi}{4} - x)$ lies within the principal value branch of $\tan^{-1} x$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

• Given: $-\frac{\pi}{4} < x < \frac{3\pi}{4}$
• Multiply by $-1$: $-\frac{3\pi}{4} < -x < \frac{\pi}{4}$
• Add $\frac{\pi}{4}$: $-\frac{2\pi}{4} < \frac{\pi}{4} - x < \frac{2\pi}{4}$
• Result: $-\frac{\pi}{2} < \frac{\pi}{4} - x < \frac{\pi}{2}$

[By the property of inverse trigonometric functions, $\tan^{-1}(\tan \theta) = \theta$ if $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$].

Step 4: Geometric Representation of the Identity

Final Simplification

Since the argument falls within the required interval, the expression simplifies directly to:

$$\tan^{-1} \left( \tan \left( \frac{\pi}{4} - x \right) \right) = \frac{\pi}{4} - x$$

Final Solution: $\frac{\pi}{4} - x$

Step 1: Analyzing the Domain and Range Constraints

To evaluate $\tan^{-1}(\tan(x))$, we must recall that the property $\tan^{-1}(\tan(x)) = x$ holds if and only if $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, which is the principal value branch of the tangent function [Per the definition of Inverse Trigonometric Functions].

In this problem, $x = \frac{3\pi}{4}$. Since $\frac{3\pi}{4} \approx 135^\circ$ and the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ corresponds to $(-90^\circ, 90^\circ)$, clearly $\frac{3\pi}{4} \notin (-\frac{\pi}{2}, \frac{\pi}{2})$.

Step 2: Utilizing Trigonometric Periodicity and Identity

Since the tangent function has a period of $\pi$, we use the identity $\tan(\pi - \theta) = -\tan(\theta)$. Furthermore, noting that $\tan(\theta) = \tan(\theta + n\pi)$, we look to simplify the expression using the supplementary angle property: $\tan(\pi - \theta) = -\tan(\theta)$. Alternatively, we use the specific identity:

$\tan(\frac{3\pi}{4}) = \tan(\pi - \frac{\pi}{4})$

Using the quadrant rule [Per the CAST rule for trigonometric functions in the second quadrant], we know that $\tan(\pi - \theta) = -\tan(\theta)$. Thus:

$\tan(\frac{3\pi}{4}) = -\tan(\frac{\pi}{4}) = -1$

Step 3: Finding the Principal Value

We now substitute the value back into the inverse function:

$\tan^{-1}(\tan(\frac{3\pi}{4})) = \tan^{-1}(-1)$

By definition, $\tan^{-1}(-1) = y$ implies $\tan(y) = -1$, where $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $\tan(\frac{\pi}{4}) = 1$, and the tangent function is an odd function [Per the property $\tan(-x) = -\tan(x)$], we have:

$\tan(-\frac{\pi}{4}) = -1$

Therefore, $y = -\frac{\pi}{4}$.

Visual Representation of the Transformation

Summary of Calculation

Expression	Value
Original Argument	$\frac{3\pi}{4}$
Equivalent Tangent Value	$-1$
Principal Value Result	$-\frac{\pi}{4}$

Final Solution: -\frac{\pi}{4}

Step 1: Decomposition of the Expression

The given expression is $\tan(\sin^{-1}(\frac{3}{5}) + \cot^{-1}(\frac{3}{2}))$. To solve this, we must express both inverse trigonometric functions in terms of $\tan^{-1}(x)$.

Let $\alpha = \sin^{-1}(\frac{3}{5})$ and $\beta = \cot^{-1}(\frac{3}{2})$. The expression becomes $\tan(\alpha + \beta)$.

Step 2: Converting Inverse Functions to $\tan^{-1}$

For $\alpha = \sin^{-1}(\frac{3}{5})$:

• $\sin(\alpha) = \frac{3}{5}$ [Definition of inverse sine].
• Since $\sin(\alpha) = \frac{\text{Opposite}}{\text{Hypotenuse}}$, in a right-angled triangle, Opposite = 3, Hypotenuse = 5.
• Adjacent side = $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$ [By Pythagorean Theorem].
• $\tan(\alpha) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4}$.
• Therefore, $\alpha = \tan^{-1}(\frac{3}{4})$.

For $\beta = \cot^{-1}(\frac{3}{2})$:

• $\cot(\beta) = \frac{3}{2}$ [Definition of inverse cotangent].
• Using the identity $\tan^{-1}(x) = \cot^{-1}(\frac{1}{x})$ for $x > 0$:
• $\beta = \tan^{-1}(\frac{1}{3/2}) = \tan^{-1}(\frac{2}{3})$.

Step 3: Applying the Tangent Sum Identity

Now, the expression is $\tan(\tan^{-1}(\frac{3}{4}) + \tan^{-1}(\frac{2}{3}))$.

We use the identity: $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, provided $xy < 1$ [Summation property of inverse tangents].

Here, $x = \frac{3}{4}$ and $y = \frac{2}{3}$.

$xy = (\frac{3}{4}) \times (\frac{2}{3}) = \frac{6}{12} = \frac{1}{2}$. Since $\frac{1}{2} < 1$, the formula is valid.

Step 4: Final Algebraic Computation

Let $\theta = \tan^{-1}(\frac{3}{4}) + \tan^{-1}(\frac{2}{3})$.

$\theta = \tan^{-1}\left( \frac{\frac{3}{4} + \frac{2}{3}}{1 - (\frac{3}{4} \times \frac{2}{3})} \right)$

$\theta = \tan^{-1}\left( \frac{\frac{9+8}{12}}{1 - \frac{6}{12}} \right)$

$\theta = \tan^{-1}\left( \frac{\frac{17}{12}}{\frac{6}{12}} \right) = \tan^{-1}\left( \frac{17}{6} \right)$

The original expression is $\tan(\theta) = \tan(\tan^{-1}(\frac{17}{6}))$.

By definition, $\tan(\tan^{-1}(z)) = z$ for all $z \in \mathbb{R}$.

Final Solution: \frac{17}{6}

Given Variables & Initial Setup

We are required to prove the identity: $3\sin^{-1} x = \sin^{-1} (3x - 4x^3)$ for $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$.

Step 1: Substitution and Domain Mapping

Let $\sin^{-1} x = \theta$. By the definition of the inverse trigonometric function, this implies $x = \sin \theta$. [Per the definition of the range of $\sin^{-1} x$, $\theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$].

Given the constraint $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$, we verify the range of $\theta$:

• If $x = -\frac{1}{2}$, $\theta = \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$.
• If $x = \frac{1}{2}$, $\theta = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$.
• Thus, $\theta \in \left[ -\frac{\pi}{6}, \frac{\pi}{6} \right]$.

Step 2: Applying the Triple Angle Identity

Recall the trigonometric identity for $\sin(3\theta)$:

$\sin(3\theta) = 3\sin \theta - 4\sin^3 \theta$ [Per the Multiple Angle Formula for sine].

Substitute $x = \sin \theta$ into the right-hand side (RHS) of the equation:

$\text{RHS} = \sin^{-1}(3x - 4x^3)$

$\text{RHS} = \sin^{-1}(3\sin \theta - 4\sin^3 \theta)$

$\text{RHS} = \sin^{-1}(\sin 3\theta)$

Step 3: Justifying the Inverse Operation

We know that $\sin^{-1}(\sin \alpha) = \alpha$ if $\alpha \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Since $\theta \in \left[ -\frac{\pi}{6}, \frac{\pi}{6} \right]$, it follows that $3\theta \in \left[ -\frac{3\pi}{6}, \frac{3\pi}{6} \right]$, which simplifies to $3\theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Therefore, $\sin^{-1}(\sin 3\theta) = 3\theta$.

Step 4: Final Substitution

From Step 1, we established that $\theta = \sin^{-1} x$.

Substituting this back into our result:

$3\theta = 3\sin^{-1} x$

Since the RHS yielded $3\theta$ and the LHS is $3\sin^{-1} x$, we have successfully demonstrated that:

$\text{LHS} = \text{RHS} = 3\sin^{-1} x$

Final Solution: $3\sin^{-1} x = \sin^{-1} (3x - 4x^3)$ is proved for $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$.

Step 1: Evaluation of the Inverse Trigonometric Component

To solve the expression $\sin(\frac{\pi}{3} - \sin^{-1}(\frac{-1}{2}))$, we must first resolve the inner inverse sine function, $\sin^{-1}(\frac{-1}{2})$.

Let $\theta = \sin^{-1}(\frac{-1}{2})$. By the definition of the range of the principal value branch of $\sin^{-1}(x)$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we have:

$\sin(\theta) = -\frac{1}{2}$ [By the definition of inverse trigonometric functions]

Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$, and given the odd function property $\sin(-\theta) = -\sin(\theta)$, it follows that:

$\sin(-\frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}$

Thus, $\theta = -\frac{\pi}{6}$, which lies within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Step 2: Substitution and Simplification of the Argument

Substitute the value of $\theta$ back into the original expression:

Expression $= \sin(\frac{\pi}{3} - (-\frac{\pi}{6}))$

$= \sin(\frac{\pi}{3} + \frac{\pi}{6})$

To add these fractions, find the common denominator, which is $6$:

$= \sin(\frac{2\pi}{6} + \frac{\pi}{6})$

$= \sin(\frac{3\pi}{6})$

$= \sin(\frac{\pi}{2})$

Step 3: Final Trigonometric Evaluation

The trigonometric value of $\sin(\frac{\pi}{2})$ is a standard unit circle identity:

$\sin(\frac{\pi}{2}) = 1$ [By the trigonometric ratio of standard angles]

Visual Representation of the Argument

Final Solution: 1

Step 1: Establishing the Domain and Range Constraints

To evaluate the expression $\cos^{-1}(\cos(x))$, we must recall that the identity $\cos^{-1}(\cos(x)) = x$ holds true if and only if $x$ lies within the principal value branch of the cosine function, which is $[0, \pi]$.

Given the input angle $\theta = \frac{7\pi}{6}$, we observe:

• $\frac{7\pi}{6} = 210^\circ$
• The interval $[0, \pi]$ corresponds to $[0^\circ, 180^\circ]$.

Since $\frac{7\pi}{6} > \pi$, the identity $\cos^{-1}(\cos(x)) = x$ cannot be applied directly [Per the definition of Inverse Trigonometric principal values].

Step 2: Reducing the Angle using Trigonometric Periodicity

We utilize the property that the cosine function is periodic with period $2\pi$, specifically $\cos(2\pi - \theta) = \cos(\theta)$. We aim to express $\frac{7\pi}{6}$ in a form that falls within $[0, \pi]$ while maintaining the equivalence of the cosine value.

Applying the reduction formula:

$\cos\left(\frac{7\pi}{6}\right) = \cos\left(2\pi - \frac{7\pi}{6}\right)$

$\cos\left(\frac{7\pi}{6}\right) = \cos\left(\frac{12\pi - 7\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right)$

[By the Quadrant property: $\cos(2\pi - \theta) = \cos(\theta)$ in the fourth quadrant where cosine is positive].

Step 3: Verification of the Principal Value

We now check if $\frac{5\pi}{6}$ resides within the principal range $[0, \pi]$.

$0 \leq \frac{5\pi}{6} \leq \pi$ is satisfied because $0 \leq 0.833\pi \leq 1\pi$.

Therefore, we can now apply the identity $\cos^{-1}(\cos(x)) = x$ for $x = \frac{5\pi}{6}$.

Step 4: Graphical Representation of the Reduction

Step 5: Final Evaluation

$\cos^{-1}\left(\cos\left(\frac{7\pi}{6}\right)\right) = \cos^{-1}\left(\cos\left(\frac{5\pi}{6}\right)\right)$

Given that $\frac{5\pi}{6} \in [0, \pi]$, we conclude:

$\cos^{-1}\left(\cos\left(\frac{5\pi}{6}\right)\right) = \frac{5\pi}{6}$

Final Solution: \frac{5\pi}{6}

Step 1: Establishing the Substitution

To prove the identity $3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$, we begin by substituting $x$ with a trigonometric function that simplifies the expression on the right-hand side. Let:

$x = \cos \theta$

Consequently, by the definition of the inverse cosine function, we have:

$\theta = \cos^{-1} x$ [Domain constraint: $x \in [\frac{1}{2}, 1]$ implies $\theta \in [0, \frac{\pi}{3}]$]

Step 2: Utilizing the Triple Angle Identity

Substitute $x = \cos \theta$ into the right-hand side (RHS) of the equation:

RHS $= \cos^{-1} (4\cos^3 \theta - 3\cos \theta)$

Recall the standard trigonometric identity for the cosine of a triple angle [Per De Moivre’s Theorem/Trigonometric Identity]:

$\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$

Substituting this identity into our expression, we obtain:

RHS $= \cos^{-1} (\cos 3\theta)$

Step 3: Verification of the Principal Value Branch

The expression $\cos^{-1} (\cos 3\theta)$ simplifies to $3\theta$ if and only if $3\theta$ lies within the principal value branch of the inverse cosine function, which is $[0, \pi]$.

• Given $x \in [\frac{1}{2}, 1]$
• $\cos \theta \in [\frac{1}{2}, 1]$
• $\theta \in [0, \frac{\pi}{3}]$
• Multiplying by 3, we get $3\theta \in [0, \pi]$

Since $3\theta \in [0, \pi]$, the property $\cos^{-1}(\cos \alpha) = \alpha$ holds for $\alpha = 3\theta$ [By the property of inverse trigonometric functions].

Step 4: Final Derivation

RHS $= 3\theta$

Substitute $\theta = \cos^{-1} x$ back into the equation:

RHS $= 3\cos^{-1} x$

Therefore, LHS = RHS.

Final Solution: $3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$ is proven for $x \in [\frac{1}{2}, 1]$.

Step 1: Evaluation of the innermost trigonometric term

We are given the expression: $f(x) = \tan^{-1} [2\cos(2\sin^{-1}(\frac{1}{2}))]$.

First, consider the term $\sin^{-1}(\frac{1}{2})$. We know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$. Since the principal value branch of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we have:

$\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ [By definition of principal value branch]

Step 2: Substitution and simplification of the argument

Substitute the value obtained in Step 1 back into the expression:

$2\sin^{-1}(\frac{1}{2}) = 2 \times (\frac{\pi}{6}) = \frac{\pi}{3}$ [Algebraic simplification]

Now, the expression transforms into:

$\tan^{-1} [2\cos(\frac{\pi}{3})]$

Step 3: Evaluation of the cosine function

Evaluate $\cos(\frac{\pi}{3})$. We know from the unit circle values that $\cos(\frac{\pi}{3}) = \frac{1}{2}$.

Substitute this into the expression:

$\tan^{-1} [2 \times \frac{1}{2}]$

$\tan^{-1} [1]$ [Arithmetic multiplication]

Step 4: Final inverse trigonometric evaluation

We seek the principal value of $\tan^{-1}(1)$. Let $\theta = \tan^{-1}(1)$, which implies $\tan(\theta) = 1$.

The principal value branch of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $\tan(\frac{\pi}{4}) = 1$, we conclude:

$\theta = \frac{\pi}{4}$

Final Solution: $\frac{\pi}{4}$

Step 1: Substitution and Variable Transformation

To simplify the expression $f(x) = \tan^{-1} \left( \frac{3a^2x - x^3}{a^3 - 3ax^2} \right)$, we recognize the structural similarity between the argument of the function and the expansion formula for $\tan(3\theta)$.

Let $x = a \tan \theta$.
Then, $\frac{x}{a} = \tan \theta \implies \theta = \tan^{-1} \left( \frac{x}{a} \right)$.

Given the domain constraint $\frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$, we divide by $a$ ($a > 0$):
$\frac{-1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$
$\tan \left( -\frac{\pi}{6} \right) < \tan \theta < \tan \left( \frac{\pi}{6} \right)$
Thus, $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$. [Constraint for validity of inverse function].

Step 2: Substitution into the Expression

Substitute $x = a \tan \theta$ into the function:

$f(\theta) = \tan^{-1} \left( \frac{3a^2(a \tan \theta) - (a \tan \theta)^3}{a^3 - 3a(a \tan \theta)^2} \right)$

Factor out $a^3$ from the numerator and denominator:

$f(\theta) = \tan^{-1} \left( \frac{a^3(3 \tan \theta - \tan^3 \theta)}{a^3(1 - 3 \tan^2 \theta)} \right)$ $f(\theta) = \tan^{-1} \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right)$

Step 3: Applying Trigonometric Identities

By the triple angle identity for tangent:
$\tan(3\theta) = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$ [Trigonometric Identity].

Therefore, the expression simplifies to:

$f(\theta) = \tan^{-1} (\tan 3\theta)$

Step 4: Range Analysis and Final Simplification

Since $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$, we multiply by $3$:
$3 \times \left( -\frac{\pi}{6} \right) < 3\theta < 3 \times \left( \frac{\pi}{6} \right)$
$-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$.

Because $3\theta$ falls within the principal branch of the inverse tangent function $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have:

$\tan^{-1}(\tan 3\theta) = 3\theta$

Substitute $\theta = \tan^{-1} \left( \frac{x}{a} \right)$ back into the result:

$f(x) = 3 \tan^{-1} \left( \frac{x}{a} \right)$

Final Solution: 3 \tan^{-1} \left( \frac{x}{a} \right)

Step 1: Application of Substitution Identities

To simplify the expression $\tan \frac{1}{2}[\sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2}]$, we utilize the standard trigonometric substitution methods [Per the substitution $x = \tan \theta$].

Recall the standard inverse trigonometric identities:

• $\sin^{-1} \left( \frac{2x}{1+x^2} \right) = 2 \tan^{-1} x$, provided $|x| \le 1$
• $\cos^{-1} \left( \frac{1-y^2}{1+y^2} \right) = 2 \tan^{-1} y$, provided $y \ge 0$

Step 2: Substitution into the Original Expression

Substitute the derived identities into the expression:

$\text{Expression} = \tan \frac{1}{2} \left[ 2 \tan^{-1} x + 2 \tan^{-1} y \right]$

Distribute the $\frac{1}{2}$ across the brackets:

$\text{Expression} = \tan \left( \tan^{-1} x + \tan^{-1} y \right)$

Step 3: Application of Inverse Tangent Addition Formula

Utilize the identity for the sum of inverse tangents [Per the Addition Formula for Inverse Trigonometric Functions]:

$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$

Since the problem states $xy < 1$, the condition for the standard addition formula is satisfied. Substituting this into our expression:

$\text{Expression} = \tan \left( \tan^{-1} \left( \frac{x+y}{1-xy} \right) \right)$

Step 4: Simplification via Property of Inverse Functions

Given the property $\tan(\tan^{-1} \theta) = \theta$ for $\theta \in \mathbb{R}$, we simplify the expression directly:

$\text{Expression} = \frac{x+y}{1-xy}$

Visual Representation of the Domain Mapping

Final Solution: \frac{x+y}{1-xy}

Step 1: Substitution Strategy

To simplify the expression $f(x) = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right)$, we analyze the internal radical term $\sqrt{1+x^2}$. This structure suggests the trigonometric identity $1 + \tan^2 \theta = \sec^2 \theta$ [Per the Pythagorean Identity].

Let $x = \tan \theta$. Consequently, $\theta = \tan^{-1} x$.

Step 2: Substitution and Algebraic Simplification

Substitute $x = \tan \theta$ into the original function:

$f(x) = \tan^{-1} \left( \frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta} \right)$

Since $1 + \tan^2 \theta = \sec^2 \theta$, the expression becomes:

$f(x) = \tan^{-1} \left( \frac{\sqrt{\sec^2 \theta}-1}{\tan \theta} \right) = \tan^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right)$ [Since $\sqrt{\sec^2 \theta} = |\sec \theta|$, and for principal values, we assume $\sec \theta > 0$].

Step 3: Conversion to Sine and Cosine

Using basic trigonometric definitions $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

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

Step 4: Applying Half-Angle Identities

Utilizing the power-reduction/half-angle identities [By Double Angle Formulae]:

• $1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right)$
• $\sin \theta = 2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)$

Substituting these into the expression:

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

Step 5: Final Evaluation

Using the property $\tan^{-1}(\tan \alpha) = \alpha$ for $\alpha \in (-\pi/2, \pi/2)$ [By Inverse Trigonometric Principle]:

$f(x) = \frac{\theta}{2}$

Substituting back $\theta = \tan^{-1} x$:

$f(x) = \frac{1}{2} \tan^{-1} x$

Final Solution: $\frac{1}{2} \tan^{-1} x$

Step 1: Analyzing the Domain and Range

The principal value branch of the inverse sine function, $\sin^{-1}(x)$, is defined on the closed interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. [Per the definition of principal value branches of inverse trigonometric functions].

Given the expression $\sin^{-1}(\sin(\frac{2\pi}{3}))$, we observe that the argument $\frac{2\pi}{3}$ lies outside the principal interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ because $\frac{2\pi}{3} \approx 2.094$ radians and $\frac{\pi}{2} \approx 1.571$ radians. Since $\frac{2\pi}{3} > \frac{\pi}{2}$, we cannot directly apply the property $\sin^{-1}(\sin \theta) = \theta$.

Step 2: Utilizing Trigonometric Identity

We use the supplementary angle property of the sine function: $\sin(\pi - \theta) = \sin(\theta)$. [Per the Allied Angle Theorems].

Applying this to the angle $\frac{2\pi}{3}$:

$\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right)$

$\sin\left(\frac{2\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)$

Step 3: Evaluating the Inverse Function

Substitute the simplified value back into the original expression:

$\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right) = \sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right)$

Since $\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, the condition for the identity $\sin^{-1}(\sin x) = x$ is now satisfied. [By the Principle of Inverse Functions].

$\sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right) = \frac{\pi}{3}$

Step 4: Geometric Representation of the Argument Mapping

The SVG above demonstrates how the angle $\frac{2\pi}{3}$ (in Quadrant II) shares the same vertical projection as the angle $\frac{\pi}{3}$ (in Quadrant I) on the unit circle, confirming the equality of their sine values.

Final Solution: \frac{\pi}{3}

Step 1: Evaluation of the First Term $\tan^{-1}(\sqrt{3})$

To evaluate $\tan^{-1}(\sqrt{3})$, let $\theta_1 = \tan^{-1}(\sqrt{3})$. This implies $\tan(\theta_1) = \sqrt{3}$.

The principal value branch of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $\tan(\frac{\pi}{3}) = \sqrt{3}$, we have:

$\theta_1 = \frac{\pi}{3}$ [Per the definition of Principal Value Branches]

Step 2: Evaluation of the Second Term $\cot^{-1}(-\sqrt{3})$

To evaluate $\cot^{-1}(-\sqrt{3})$, let $\theta_2 = \cot^{-1}(-\sqrt{3})$. This implies $\cot(\theta_2) = -\sqrt{3}$.

The principal value branch of $\cot^{-1}(x)$ is $(0, \pi)$. [Per the definition of Principal Value Branches for inverse cotangent functions]

Using the identity $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$:

$\theta_2 = \pi - \cot^{-1}(\sqrt{3})$

Since $\cot(\frac{\pi}{6}) = \sqrt{3}$, it follows that $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$.

$\theta_2 = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Step 3: Calculating the Final Difference

Now, substitute the evaluated values into the original expression:

Expression $= \tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})$

Expression $= \frac{\pi}{3} - \frac{5\pi}{6}$

To perform the subtraction, find a common denominator:

Expression $= \frac{2\pi}{6} - \frac{5\pi}{6}$

Expression $= -\frac{3\pi}{6}$

Expression $= -\frac{\pi}{2}$

Visual Representation of Angles in the Unit Circle

Final Solution: -\frac{\pi}{2}