Step 1: Establishing the Domain and Substitution

Given the equation: $\sin^{-1}(1 - x) - 2\sin^{-1} x = \frac{\pi}{2}$

We restrict the variable $x$ based on the domain of the inverse sine function, $\sin^{-1} \theta$, where $\theta \in [-1, 1]$. Thus:

• $1 - x \in [-1, 1] \implies x \in [0, 2]$
• $x \in [-1, 1]$

Combining these, we must have $x \in [0, 1]$. [Per the definition of the domain of $\sin^{-1}$].

Step 2: Rearranging the Equation

Isolate the term $\sin^{-1}(1-x)$: $$\sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1} x$$ Apply the sine function to both sides: $$1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1} x\right)$$ [Using the trigonometric identity $\sin(\frac{\pi}{2} + \theta) = \cos \theta$]: $$1 - x = \cos(2\sin^{-1} x)$$

Step 3: Algebraic Substitution and Expansion

Let $\theta = \sin^{-1} x$, which implies $\sin \theta = x$. Substituting this into the identity $\cos(2\theta) = 1 - 2\sin^2 \theta$: $$1 - x = 1 - 2\sin^2(\sin^{-1} x)$$ $$1 - x = 1 - 2x^2$$ [By the property $\sin(\sin^{-1} x) = x$ for $x \in [-1, 1]$].

Step 4: Solving the Quadratic Equation

Subtract 1 from both sides: $$-x = -2x^2$$ Rearrange to standard quadratic form: $$2x^2 - x = 0$$ Factor the expression: $$x(2x - 1) = 0$$ This yields two potential roots:

• $x_1 = 0$
• $x_2 = \frac{1}{2}$

Step 5: Verification of Roots

We test both roots against the original equation to ensure they do not produce extraneous solutions:

Value	LHS Calculation	Result
$x = 0$	$\sin^{-1}(1-0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \frac{\pi}{2}$	Valid ($\frac{\pi}{2} = \frac{\pi}{2}$)
$x = \frac{1}{2}$	$\sin^{-1}(1 - \frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \sin^{-1}(\frac{1}{2}) - 2(\frac{\pi}{6}) = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6}$	Invalid ($-\frac{\pi}{6} \neq \frac{\pi}{2}$)

Visual Representation of the Functions

Final Solution: x = 0

Step 1: Establishing the Theoretical Framework

To find the principal value of $y = \cot^{-1}(\sqrt{3})$, we must identify an angle $y$ such that:

• $\cot(y) = \sqrt{3}$
• $y$ falls within the principal value branch of the cotangent function.

[Per the definition of Inverse Trigonometric Functions, the range of the principal value branch of $\cot^{-1}(x)$ is $(0, \pi)$].

Step 2: Conversion to Reciprocal Trigonometric Identity

Since $\cot(y) = \frac{1}{\tan(y)}$, we can rewrite the equation as follows:

$\cot(y) = \sqrt{3} \implies \frac{1}{\tan(y)} = \sqrt{3}$ [By the Reciprocal Identity of Trigonometric Functions]

$\tan(y) = \frac{1}{\sqrt{3}}$

Step 3: Determining the Angle via Trigonometric Table Values

We know from the standard values of trigonometric functions that:

$\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$

Since $\frac{\pi}{6}$ lies within the interval $(0, \pi)$, it satisfies the principal value requirement.

Step 4: Verification of the Principal Value

The value $\frac{\pi}{6}$ corresponds to $30^\circ$. We confirm:

$\cot\left(\frac{\pi}{6}\right) = \frac{\cos(\pi/6)}{\sin(\pi/6)} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$

Since $0 < \frac{\pi}{6} < \pi$, this value is the principal value.

Visual Representation of the Principal Angle

Step 5: Final Evaluation

Given $\cot(y) = \sqrt{3}$, we set $y = \cot^{-1}(\sqrt{3})$.

$y = \frac{\pi}{6}$

Final Solution: The principal value of $\cot^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.

Step 1: Identifying the Definition and Domain

The principal value branch of the function $f(x) = \sin^{-1}(x)$ is defined as the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$. We seek a value $y$ such that:

$\sin(y) = -\frac{1}{2}$, where $y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ [Per the definition of principal value branches of inverse trigonometric functions].

Step 2: Evaluating the Trigonometric Ratio

We know from standard trigonometric table values that $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Utilizing the trigonometric identity for negative angles, specifically $\sin(-\theta) = -\sin(\theta)$ [By the Odd Function property of the Sine function], we apply the substitution:

$-\sin\left(\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{6}\right)$

Therefore, $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.

Step 3: Verification of the Principal Range

We must confirm that the identified angle $y = -\frac{\pi}{6}$ satisfies the condition $y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Since $-\frac{\pi}{2} \leq -\frac{\pi}{6} \leq \frac{\pi}{2}$ is a true mathematical statement, $-\frac{\pi}{6}$ lies within the principal value branch.

Step 4: Visualizing the Angle on the Unit Circle

Step 5: Final Derivation

Given $\sin(y) = -\frac{1}{2}$, then:

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

$y = -\frac{\pi}{6}$

Final Solution: The principal value of $\sin^{-1} (-\frac{1}{2})$ is $-\frac{\pi}{6}$.

Step 1: Analyzing the Domain of the Inverse Cosine Function

The inverse cosine function, $\cos^{-1}(x)$, is defined such that its principal value branch is $[0, \pi]$. Therefore, for any value $y$, the expression $\cos^{-1}(\cos(y)) = y$ holds true if and only if $y \in [0, \pi]$. [By the definition of the Principal Value Branch of Inverse Trigonometric Functions].

Step 2: Evaluating the Input Angle

The given argument is $y = \frac{13\pi}{6}$. We observe that:

$\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$

Since $\frac{13\pi}{6} > \pi$, it falls outside the principal value branch $[0, \pi]$. Consequently, we must utilize the periodicity and symmetry properties of the cosine function to reduce the angle.

Step 3: Applying Trigonometric Reduction Formulas

Using the periodicity property $\cos(2n\pi + \theta) = \cos(\theta)$ where $n \in \mathbb{Z}$ [Per the Periodic Property of Trigonometric Functions]:

$\cos\left(\frac{13\pi}{6}\right) = \cos\left(2\pi + \frac{\pi}{6}\right)$

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

Step 4: Visualizing the Transformation

Step 5: Final Calculation

Now, substitute the simplified angle back into the inverse function:

$\cos^{-1}\left(\cos\left(\frac{13\pi}{6}\right)\right) = \cos^{-1}\left(\cos\left(\frac{\pi}{6}\right)\right)$

Since $\frac{\pi}{6} \in [0, \pi]$, we apply the identity $\cos^{-1}(\cos(\theta)) = \theta$ [By the Principal Value Property]:

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

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

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: Establishing the Substitution

To simplify the expression, we begin by substituting a trigonometric variable for $\sqrt{x}$. Let $\theta = \tan^{-1} \sqrt{x}$.
This implies:
$\tan \theta = \sqrt{x}$
Given the domain $x \in [0, 1]$, it follows that $\sqrt{x} \in [0, 1]$, which restricts $\theta$ to the interval $[0, \frac{\pi}{4}]$. [Per the definition of the range of $\tan^{-1} \theta$]

Step 2: Manipulating the Right-Hand Side (RHS)

Consider the RHS: $\frac{1}{2} \cos^{-1} \left( \frac{1-x}{1+x} \right)$.
Since $\tan \theta = \sqrt{x}$, then $x = \tan^2 \theta$. Substituting this into the RHS:
$\text{RHS} = \frac{1}{2} \cos^{-1} \left( \frac{1-\tan^2 \theta}{1+\tan^2 \theta} \right)$

Step 3: Applying Trigonometric Identities

We utilize the double-angle formula for cosine in terms of tangent:
$\cos(2\theta) = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}$ [By the Double Angle Identity for $\cos(2\theta)$]

Substituting this identity into our expression:
$\text{RHS} = \frac{1}{2} \cos^{-1} (\cos 2\theta)$

Step 4: Simplifying the Inverse Function

Since $\theta \in [0, \frac{\pi}{4}]$, it follows that $2\theta \in [0, \frac{\pi}{2}]$.
The range of the principal value branch of $\cos^{-1}$ is $[0, \pi]$.
Since $2\theta$ lies within this range, we can apply the identity $\cos^{-1}(\cos \alpha) = \alpha$ for $\alpha \in [0, \pi]$. [By the Property of Inverse Trigonometric Functions]

$\text{RHS} = \frac{1}{2} (2\theta) = \theta$

Step 5: Visual Representation

Step 6: Conclusion of Equality

From Step 1, we defined $\theta = \tan^{-1} \sqrt{x}$.
From Step 4, we simplified the RHS to $\theta$.
Therefore:
$\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} \left( \frac{1-x}{1+x} \right)$

Final Solution: The identity is proven as $\text{LHS} = \text{RHS} = \theta$, given the domain constraint $x \in [0, 1]$.

Step 1: Simplifying the Radicals

To simplify expressions involving $\sqrt{1 \pm \sin x}$, we utilize the fundamental trigonometric identities: $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin x = 2\sin(x/2)\cos(x/2)$. [Per the Pythagorean Identity and Double Angle Formula].

We can rewrite the term inside the radical as:

$1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2$

$1 - \sin x = \cos^2(x/2) + \sin^2(x/2) - 2\sin(x/2)\cos(x/2) = (\cos(x/2) - \sin(x/2))^2$

Step 2: Applying Square Roots

Given $x \in (0, \pi/4)$, we know that $x/2 \in (0, \pi/8)$. In this interval, $\cos(x/2) > \sin(x/2) > 0$.

Therefore:

$\sqrt{1 + \sin x} = \sqrt{(\cos(x/2) + \sin(x/2))^2} = \cos(x/2) + \sin(x/2)$

$\sqrt{1 - \sin x} = \sqrt{(\cos(x/2) - \sin(x/2))^2} = \cos(x/2) - \sin(x/2)$

Step 3: Substitution into the Expression

Let the argument of the cotangent inverse be $A$. Substituting the results from Step 2:

$A = \frac{(\cos(x/2) + \sin(x/2)) + (\cos(x/2) - \sin(x/2))}{(\cos(x/2) + \sin(x/2)) - (\cos(x/2) - \sin(x/2))}$

Simplifying the numerator and denominator:

$A = \frac{2\cos(x/2)}{2\sin(x/2)} = \frac{\cos(x/2)}{\sin(x/2)} = \cot(x/2)$

Step 4: Final Inverse Operation

The expression now becomes:

$\cot^{-1} (\cot(x/2))$

Since $x/2 \in (0, \pi/8)$, and the principal value branch of $\cot^{-1}$ is $(0, \pi)$, we apply the property $\cot^{-1}(\cot \theta) = \theta$ for $\theta \in (0, \pi)$. [Per the definition of Inverse Trigonometric Functions].

Therefore, $\cot^{-1}(\cot(x/2)) = x/2$.

Geometric Representation of Identity

Final Solution: $\cot^{-1} \left( \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} \right) = \frac{x}{2}$

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: Analyzing the Domain of the Principal Value Branch

The function $f(x) = \tan^{-1}(\tan x)$ is defined such that $f(x) = x$ only if $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. [By definition of the Principal Value Branch of Inverse Tangent].

In this problem, the input value is $\theta = \frac{7\pi}{6}$. Comparing this to the domain:

• $\frac{7\pi}{6} \approx 1.166\pi$
• Upper limit of range: $\frac{\pi}{2} = 0.5\pi$
• Since $\frac{7\pi}{6} > \frac{\pi}{2}$, we cannot directly equate the expression to $\frac{7\pi}{6}$.

Step 2: Applying Trigonometric Periodicity and Reduction Formulae

The tangent function has a period of $\pi$, meaning $\tan(\theta) = \tan(\theta - n\pi)$ for any integer $n$. To bring the argument into the required range $(-\frac{\pi}{2}, \frac{\pi}{2})$, we perform the following reduction:

$\tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right)$

[By the Quadrant Reduction Rule: $\tan(\pi + \theta) = \tan \theta$].

Therefore, $\tan\left(\frac{7\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right)$.

Step 3: Evaluating the Inverse Function

Now, substitute the simplified expression back into the inverse function:

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

Since $\frac{\pi}{6} \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we can apply the identity $\tan^{-1}(\tan x) = x$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. [By property of inverse trigonometric functions].

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

Visual Representation of the Periodic Reduction

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

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: Theoretical Foundation of Principal Value Branch

The inverse secant function, denoted as $\sec^{-1}(x)$, is defined such that its principal value branch lies within the interval $[0, \pi] \setminus \{\frac{\pi}{2}\}$. For a given value $x$, the principal value $y$ must satisfy the condition:

$\sec(y) = x$, where $y \in [0, \pi], y \neq \frac{\pi}{2}$ [Per the definition of the range of the inverse secant function].

Step 2: Mathematical Setup and Transformation

Given the equation: $\sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = y$

By the definition of the inverse trigonometric function, we convert the expression into its equivalent trigonometric form:

$\sec(y) = \frac{2}{\sqrt{3}}$

Since $\sec(y) = \frac{1}{\cos(y)}$, we apply the reciprocal identity:

$\cos(y) = \frac{1}{\sec(y)} = \frac{1}{\frac{2}{\sqrt{3}}} = \frac{\sqrt{3}}{2}$

Step 3: Determining the Principal Value

To solve $\cos(y) = \frac{\sqrt{3}}{2}$ for $y$ within the principal interval $[0, \pi]$:

• We know from standard trigonometric table values that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$. [By the unit circle definition of cosine].
• Since $\frac{\pi}{6}$ lies within the interval $[0, \pi]$, it satisfies the condition of the principal value branch.

Step 4: Geometric Visualization

Step 5: Verification of Constraints

Check: Does $y = \frac{\pi}{6}$ satisfy $y \in [0, \pi] \setminus \{\frac{\pi}{2}\}$?

Since $0 \leq \frac{\pi}{6} \leq \pi$ and $\frac{\pi}{6} \neq \frac{\pi}{2}$, the value is valid. [Per the domain and range constraints of $\sec^{-1}(x)$].

Final Solution: \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = \frac{\pi}{6}

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

Let $y_1 = \tan^{-1}(\sqrt{3})$. By the definition of the inverse trigonometric function, this is equivalent to $\tan(y_1) = \sqrt{3}$, where the principal value branch of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $\tan(\frac{\pi}{3}) = \sqrt{3}$ [Trigonometric Standard Values], and $\frac{\pi}{3} \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we conclude:

$y_1 = \frac{\pi}{3}$

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

Let $y_2 = \sec^{-1}(-2)$. By definition, $\sec(y_2) = -2$, where the principal value branch of $\sec^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$.

Recall the identity $\sec(y) = \frac{1}{\cos(y)}$. Thus, $\cos(y_2) = -\frac{1}{2}$ [Reciprocal Identity].

We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$. Since the cosine function is negative in the second quadrant, we use the identity $\cos(\pi - \theta) = -\cos(\theta)$ [Allied Angle Formula]:

$\cos(\pi - \frac{\pi}{3}) = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$

Since $\frac{2\pi}{3} \in [0, \pi]$, we conclude:

$y_2 = \frac{2\pi}{3}$

Step 3: Geometrical Representation of Principal Values

Step 4: Final Calculation

Substitute the evaluated principal values into the original expression:

$\text{Result} = y_1 - y_2$

$\text{Result} = \frac{\pi}{3} - \frac{2\pi}{3}$

$\text{Result} = -\frac{\pi}{3}$

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

Step 1: Identifying the Left-Hand Side (LHS) Transformation

We begin with the expression: $2\sin^{-1} \left( \frac{3}{5} \right)$. To simplify this into a tangent form, we first express the $\sin^{-1}$ term as an angle $\theta$. Let:

$\sin^{-1} \left( \frac{3}{5} \right) = \theta \implies \sin \theta = \frac{3}{5}$ [Definition of Inverse Trigonometric Functions]

Step 2: Calculating Geometric Ratios

In a right-angled triangle where the opposite side $O = 3$ and the hypotenuse $H = 5$, we find the adjacent side $A$ using the Pythagorean theorem $A^2 + O^2 = H^2$:

$A = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$

Therefore, $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4}$.

Thus, $\sin^{-1} \left( \frac{3}{5} \right) = \tan^{-1} \left( \frac{3}{4} \right)$.

Step 3: Applying the Double Angle Formula

The original expression becomes $2\tan^{-1} \left( \frac{3}{4} \right)$. We utilize the standard identity for $2\tan^{-1} x$:

$2\tan^{-1} x = \tan^{-1} \left( \frac{2x}{1 - x^2} \right)$ [Standard Identity for Inverse Tangent]

Substituting $x = \frac{3}{4}$:

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

Step 4: Algebraic Simplification

Calculate the numerator and denominator:

• Numerator: $2 \times \frac{3}{4} = \frac{3}{2} = 1.5$
• Denominator: $1 - \frac{9}{16} = \frac{16-9}{16} = \frac{7}{16}$

Combining these into the tangent argument:

$\frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} \times \frac{16}{7} = \frac{3 \times 8}{7} = \frac{24}{7}$

Step 5: Conclusion of Proof

Since the simplification yields $\tan^{-1} \left( \frac{24}{7} \right)$, which is exactly equal to the Right-Hand Side (RHS), the identity is proven.

Final Solution: $2\sin^{-1} \left( \frac{3}{5} \right) = \tan^{-1} \left( \frac{24}{7} \right)$

Step 1: Identifying the Algebraic Transformation

The given equation is $\tan^{-1} \left( \frac{1-x}{1+x} \right) = \frac{1}{2} \tan^{-1} x$. We recognize the left-hand side as a standard identity form. Recall the formula for the difference of tangents: $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$.

Step 2: Decomposition using Trigonometric Identities

Rewrite the expression $\frac{1-x}{1+x}$ as $\frac{1-x}{1+(1)(x)}$. Applying the identity [$\tan^{-1} 1 - \tan^{-1} x = \tan^{-1} \left( \frac{1-x}{1+1 \cdot x} \right)$]:

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

Step 3: Solving the Linear Equation in terms of $\tan^{-1} x$

Substitute the principal value of $\tan^{-1}(1)$, which is $\frac{\pi}{4}$:

$\frac{\pi}{4} - \tan^{-1}(x) = \frac{1}{2} \tan^{-1}(x)$

Add $\tan^{-1}(x)$ to both sides of the equation [By the Additive Property of Equality]:

$\frac{\pi}{4} = \frac{1}{2} \tan^{-1}(x) + \tan^{-1}(x)$

$\frac{\pi}{4} = \frac{3}{2} \tan^{-1}(x)$

Step 4: Isolating $x$

Multiply by $\frac{2}{3}$ to isolate $\tan^{-1}(x)$:

$\tan^{-1}(x) = \frac{\pi}{4} \cdot \frac{2}{3} = \frac{\pi}{6}$

Take the tangent of both sides [Definition of Inverse Trigonometric Functions]:

$x = \tan\left(\frac{\pi}{6}\right)$

$x = \frac{1}{\sqrt{3}}$

Step 5: Verification of Domain Constraint

The problem provides the condition $x > 0$. Since $\frac{1}{\sqrt{3}} \approx 0.577$, the result satisfies $x > 0$ [Condition Check].

Final Solution: x = \frac{1}{\sqrt{3}}

Step 1: Theoretical Foundation of Principal Value

To find the principal value of $\csc^{-1}(x)$, we must identify the angle $\theta$ such that $\csc(\theta) = x$, where $\theta$ lies within the defined principal value branch for the inverse cosecant function. The domain of $\csc^{-1}(x)$ is $(-\infty, -1] \cup [1, \infty)$ and its range is $[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$.

Let $y = \csc^{-1}(2)$. This implies:
$\csc(y) = 2$ [By definition of inverse trigonometric functions]

Step 2: Relation to Primary Trigonometric Ratios

Since $\csc(y) = \frac{1}{\sin(y)}$, we can convert the equation into terms of sine:
$\frac{1}{\sin(y)} = 2$
$\sin(y) = \frac{1}{2}$

Step 3: Solving for the Principal Angle

We seek the angle $y \in [-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$ such that $\sin(y) = \frac{1}{2}$.
From the standard trigonometric unit circle values, we know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$.

Since $\frac{\pi}{6}$ falls within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\frac{\pi}{6} \neq 0$, it is the principal value.

Step 4: Geometric Visualization

Step 5: Logical Verification

Expression	Value
$\csc(y) = 2$	Confirmed
$y \in [-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$	$\frac{\pi}{6} \approx 0.523$ radians; Condition satisfied.

Final Solution: The principal value of $\csc^{-1}(2)$ is $\frac{\pi}{6}$.

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: Definition of the Inverse Sine Function

By definition, the inverse sine function, denoted as $\sin^{-1} x$ or $\arcsin x$, is the inverse function of the restricted sine function. For a function $f(x) = \sin x$ to be invertible, its domain must be restricted to the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ to ensure it is bijective (both injective and surjective) over the range $[-1, 1]$.

[Per the definition of Inverse Trigonometric Functions, the domain of $\sin x$ is restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ to map the range $[-1, 1]$ uniquely.]

Step 2: Establishing the Range for $y$

Given the equation $y = \sin^{-1} x$, the variable $y$ represents the principal value of the angle whose sine is $x$. According to the standard principal value branches of inverse trigonometric functions:

• The domain of $\sin^{-1} x$ is $x \in [-1, 1]$.
• The range (principal value branch) of $\sin^{-1} x$ is $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Step 3: Visual Representation of the Principal Value Branch

Step 4: Formal Conclusion of the Inequality

Since $y$ is defined as the output of the principal branch of the inverse sine function, $y$ must be greater than or equal to the lower bound and less than or equal to the upper bound of the restricted interval.

Mathematically, this is expressed as:

$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

[By the properties of the principal value branch for the inverse sine function as established in standard trigonometric identity frameworks.]

Final Solution: $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

Step 1: Identifying the Principle Value Branch

The function $f(x) = \tan^{-1}(x)$ has a defined principal value branch restricted to the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. [Per the definition of the inverse tangent function].

We are tasked with finding the principal value $y$ such that:

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

By the definition of inverse functions, this is equivalent to:

$$\tan(y) = -\sqrt{3}, \quad y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

Step 2: Applying Trigonometric Identities

We observe that $\tan(\frac{\pi}{3}) = \sqrt{3}$. Using the odd function property of the tangent function, specifically $\tan(-\theta) = -\tan(\theta)$ [Per the property of odd trigonometric functions], we can write:

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

Therefore:

$$\tan(y) = \tan\left(-\frac{\pi}{3}\right)$$

Step 3: Verification of the Domain

We must confirm that the value $-\frac{\pi}{3}$ lies within the principal value branch $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $-\frac{\pi}{2} < -\frac{\pi}{3} < \frac{\pi}{2}$, the value satisfies the criteria for the principal value branch [By the definition of the range of the inverse tangent function].

Step 4: Conclusion

Given $\tan(y) = \tan(-\frac{\pi}{3})$ and $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$, it follows that $y = -\frac{\pi}{3}$.

Final Solution: The principal value of $\tan^{-1}(-\sqrt{3})$ is $-\frac{\pi}{3}$.

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}

Step 1: Establishing the Substitution

To evaluate the expression $\sin(\tan^{-1} x)$, we begin by assigning a variable to the inverse trigonometric component. Let:

$\theta = \tan^{-1} x$

By the definition of the inverse tangent function, this implies:

$\tan \theta = x$ [Where $\theta \in (-\pi/2, \pi/2)$]

Step 2: Geometric Representation

Consider a right-angled triangle where the angle is $\theta$. Since $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{1}$, we assign:

• Opposite side = $x$
• Adjacent side = $1$

By the Pythagorean theorem ($a^2 + b^2 = c^2$), the hypotenuse $h$ is calculated as:

$h = \sqrt{x^2 + 1^2} = \sqrt{1 + x^2}$

Step 3: Evaluating the Outer Trigonometric Function

The original expression is $\sin(\theta)$. Using the ratio definition of sine in a right-angled triangle:

$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$

Substituting the values derived in Step 2:

$\sin \theta = \frac{x}{\sqrt{1 + x^2}}$

Step 4: Algebraic Verification

Using trigonometric identities, we know that $\sin \theta$ can be expressed in terms of $\tan \theta$ via the identity $\sec^2 \theta = 1 + \tan^2 \theta$:

$\cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1 + \tan^2 \theta}}$

Since $\sin \theta = \tan \theta \cdot \cos \theta$:

$\sin \theta = \tan \theta \cdot \frac{1}{\sqrt{1 + \tan^2 \theta}}$

Substituting $\tan \theta = x$:

$\sin \theta = \frac{x}{\sqrt{1 + x^2}}$ [Per the trigonometric quotient identity]

Final Solution: \frac{x}{\sqrt{1+x^2}}