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:

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

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:

Visual Representation of the Functions

Final Solution: x = 0

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: 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: 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: 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: 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}}

Step 1: Identifying the Theoretical Framework

To prove the identity $\cos^{-1} \left(\frac{4}{5}\right) + \cos^{-1} \left(\frac{12}{13}\right) = \cos^{-1} \left(\frac{33}{65}\right)$, we utilize the standard inverse trigonometric addition formula [Per the identity for $\cos^{-1} x + \cos^{-1} y$]:

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

where $x, y \in [-1, 1]$ and $x^2 + y^2 \leq 1$ [Condition for Principal Value Branch].

Step 2: Assigning Variables and Calculating Components

Let $x = \frac{4}{5}$ and $y = \frac{12}{13}$. We must compute the square roots of $(1-x^2)$ and $(1-y^2)$:

• $\sqrt{1 - x^2} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$
• $\sqrt{1 - y^2} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$

Step 3: Applying the Addition Formula

Substitute the derived values into the addition identity:

$\cos^{-1} \left( \frac{4}{5} \right) + \cos^{-1} \left( \frac{12}{13} \right) = \cos^{-1} \left( \left(\frac{4}{5} \cdot \frac{12}{13}\right) - \left(\frac{3}{5} \cdot \frac{5}{13}\right) \right)$

Step 4: Algebraic Simplification

Perform the multiplication and subtraction of the fractions within the inverse cosine function:

$= \cos^{-1} \left( \frac{48}{65} - \frac{15}{65} \right)$

$= \cos^{-1} \left( \frac{48 - 15}{65} \right)$

$= \cos^{-1} \left( \frac{33}{65} \right)$

Visual Representation of Triangle Ratios

Final Solution: $\cos^{-1} \left(\frac{4}{5}\right) + \cos^{-1} \left(\frac{12}{13}\right) = \cos^{-1} \left(\frac{33}{65}\right)$ [Q.E.D.]

Step 1: Analyzing the Right-Hand Side (RHS)

To prove the identity $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$, we first convert the individual inverse trigonometric terms on the RHS into $\tan^{-1}$ functions to utilize the addition formula: $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$.

Step 2: Conversion of Terms

• For $\sin^{-1} \frac{5}{13}$: Let $\alpha = \sin^{-1} \frac{5}{13}$. Thus, $\sin \alpha = \frac{5}{13}$. Using the Pythagorean identity $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - (\frac{5}{13})^2} = \sqrt{\frac{144}{169}} = \frac{12}{13}$. Therefore, $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{5/13}{12/13} = \frac{5}{12}$. So, $\sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}$. [Per the definition of trigonometric ratios in a right-angled triangle]
• For $\cos^{-1} \frac{3}{5}$: Let $\beta = \cos^{-1} \frac{3}{5}$. Thus, $\cos \beta = \frac{3}{5}$. Using the identity $\sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. Therefore, $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/5}{3/5} = \frac{4}{3}$. So, $\cos^{-1} \frac{3}{5} = \tan^{-1} \frac{4}{3}$.

Step 3: Applying the Addition Formula

Substitute the converted terms into the RHS expression:

$RHS = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$

Apply the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$ where $x = \frac{5}{12}$ and $y = \frac{4}{3}$:

Calculation of argument: $\frac{\frac{5}{12} + \frac{4}{3}}{1 - (\frac{5}{12} \times \frac{4}{3})}$

Numerator: $\frac{5 + 16}{12} = \frac{21}{12} = \frac{7}{4}$

Denominator: $1 - \frac{20}{36} = \frac{36 - 20}{36} = \frac{16}{36} = \frac{4}{9}$

Result of fraction: $\frac{7/4}{4/9} = \frac{7}{4} \times \frac{9}{4} = \frac{63}{16}$

Visual Representation of Triangle Conversions

Step 4: Final Conclusion

By equating the computed values, we obtain:

$\tan^{-1} \left( \frac{63}{16} \right) = \tan^{-1} \left( \frac{63}{16} \right)$

This confirms that $LHS = RHS$.

Final Solution: The identity $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$ is proven to be true.

Step 1: Conversion of Inverse Trigonometric Functions to a Common Domain

To evaluate the sum $\cos^{-1} \left( \frac{12}{13} \right) + \sin^{-1} \left( \frac{3}{5} \right)$, it is most efficient to express both terms as $\sin^{-1}$ functions. [Per the identity: $\cos^{-1}(x) = \sin^{-1}(\sqrt{1-x^2})$ for $x \in [0, 1]$].

Let $\alpha = \cos^{-1} \left( \frac{12}{13} \right)$. Then, $\cos \alpha = \frac{12}{13}$.

Using the Pythagorean identity $\sin^2 \alpha + \cos^2 \alpha = 1$:

$\sin \alpha = \sqrt{1 - \left( \frac{12}{13} \right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$.

Therefore, $\alpha = \sin^{-1} \left( \frac{5}{13} \right)$.

Step 2: Substitution and Algebraic Application

The original expression becomes:

$\sin^{-1} \left( \frac{5}{13} \right) + \sin^{-1} \left( \frac{3}{5} \right)$

We apply the addition formula for inverse sine functions: [Per the identity: $\sin^{-1} x + \sin^{-1} y = \sin^{-1} \left( x\sqrt{1-y^2} + y\sqrt{1-x^2} \right)$].

Here, $x = \frac{5}{13}$ and $y = \frac{3}{5}$.

Step 3: Execution of the Calculation

Substitute the values into the formula:

$\sin^{-1} \left( \left( \frac{5}{13} \right) \sqrt{1 - \left( \frac{3}{5} \right)^2} + \left( \frac{3}{5} \right) \sqrt{1 - \left( \frac{5}{13} \right)^2} \right)$

Calculate the terms within the square roots:

$\sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

$\sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$

Compute the products:

$\sin^{-1} \left( \left( \frac{5}{13} \cdot \frac{4}{5} \right) + \left( \frac{3}{5} \cdot \frac{12}{13} \right) \right)$

$\sin^{-1} \left( \frac{20}{65} + \frac{36}{65} \right)$

$\sin^{-1} \left( \frac{56}{65} \right)$

Step 4: Geometrical Representation

The diagram illustrates the right-angled triangles corresponding to $\cos^{-1}(12/13)$ and $\sin^{-1}(3/5)$, where the sum of angles $\alpha + \beta$ results in the angle whose sine is $56/65$.

Final Solution: $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$

Step 1: Substitution Strategy

To simplify the trigonometric expression within the inverse function, we introduce a substitution that utilizes the identity $\cos(2\theta) = 2\cos^2\theta - 1$. Let $x = \cos(2\theta)$.

Given the constraint $\frac{-1}{\sqrt{2}} \leq x \leq 1$, we find the range for $\theta$:

• If $x = 1$, $\cos(2\theta) = 1 \implies 2\theta = 0 \implies \theta = 0$.
• If $x = \frac{-1}{\sqrt{2}}$, $\cos(2\theta) = \frac{-1}{\sqrt{2}} \implies 2\theta = \frac{3\pi}{4} \implies \theta = \frac{3\pi}{8}$.

Thus, $0 \leq 2\theta \leq \frac{3\pi}{4}$, which implies $0 \leq \theta \leq \frac{3\pi}{8}$. [By substitution method for trigonometric simplification].

Step 2: Simplifying the Algebraic Expression

Substituting $x = \cos(2\theta)$ into the expression $\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}$:

$1 + \cos(2\theta) = 2\cos^2\theta$

$1 - \cos(2\theta) = 2\sin^2\theta$

Applying these identities [Using half-angle identities]:

$\frac{\sqrt{2\cos^2\theta} - \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta} + \sqrt{2\sin^2\theta}} = \frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta} = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$

Step 3: Applying Tangent Addition Formula

Divide the numerator and denominator by $\cos\theta$:

$\frac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right)$ [Using $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ where $\tan(\frac{\pi}{4}) = 1$].

Step 4: Evaluating the Inverse Function

We now evaluate $\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right)$.

Since $\theta \in [0, \frac{3\pi}{8}]$, then $(\frac{\pi}{4} - \theta) \in [\frac{\pi}{4} - \frac{3\pi}{8}, \frac{\pi}{4} - 0] = [-\frac{\pi}{8}, \frac{\pi}{4}]$.

This range lies within the principal value branch of $\tan^{-1}(x)$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Therefore, $\tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta$.

Step 5: Back-Substitution

Since $x = \cos(2\theta)$, then $2\theta = \cos^{-1}x$, which implies $\theta = \frac{1}{2}\cos^{-1}x$.

Substituting back into the expression:

$\frac{\pi}{4} - \frac{1}{2}\cos^{-1}x$.

Final Solution: \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x

Step 1: Conversion of Inverse Sine Terms to Inverse Tangent Terms

To prove the identity $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$, we first express each $\sin^{-1}$ term in the form of $\tan^{-1}$ using the trigonometric identity $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$.

For $\sin^{-1} \frac{8}{17}$:

• Let $\alpha = \sin^{-1} \frac{8}{17} \implies \sin \alpha = \frac{8}{17}$.
• Using the Pythagorean theorem, $\text{adj} = \sqrt{\text{hyp}^2 - \text{opp}^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$. [By Pythagorean Identity: $a^2 + b^2 = c^2$].
• Therefore, $\tan \alpha = \frac{8}{15} \implies \alpha = \tan^{-1} \frac{8}{15}$.

For $\sin^{-1} \frac{3}{5}$:

• Let $\beta = \sin^{-1} \frac{3}{5} \implies \sin \beta = \frac{3}{5}$.
• Using the Pythagorean theorem, $\text{adj} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.
• Therefore, $\tan \beta = \frac{3}{4} \implies \beta = \tan^{-1} \frac{3}{4}$.

Step 2: Applying the Tangent Addition Formula

We now substitute these values into the left-hand side (LHS) of the original equation: $\tan^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}$.

Using the inverse trigonometric formula: $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right)$, provided $xy < 1$. [Condition check: $(\frac{8}{15}) \times (\frac{3}{4}) = \frac{24}{60} = 0.4 < 1$].

Step 3: Algebraic Computation

Substitute $x = \frac{8}{15}$ and $y = \frac{3}{4}$ into the identity:

$\text{LHS} = \tan^{-1} \left( \frac{\frac{8}{15} + \frac{3}{4}}{1 - (\frac{8}{15} \times \frac{3}{4})} \right)$

Calculating the numerator:

Calculating the denominator:

Dividing the fractions:

Visual Representation of Triangle Constants

Final Solution: $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$ (Q.E.D.)

Step 1: Application of the Inverse Tangent Double Angle Identity

Given the equation: $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$

We apply the identity $2\tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$, valid for $|\theta| < 1$. [Per the Properties of Inverse Trigonometric Functions]

Substituting $\theta = \cos x$, we get:

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

Step 2: Algebraic Simplification

By applying the tangent function to both sides, we eliminate the inverse operators:

$\frac{2\cos x}{1-\cos^2 x} = 2\csc x$

Using the trigonometric identity $1-\cos^2 x = \sin^2 x$, we rewrite the denominator:

$\frac{2\cos x}{\sin^2 x} = 2 \cdot \frac{1}{\sin x}$ [Since $\csc x = \frac{1}{\sin x}$]

Dividing both sides by 2 and assuming $\sin x \neq 0$:

$\frac{\cos x}{\sin^2 x} = \frac{1}{\sin x}$

Step 3: Solving for $x$

Multiply both sides by $\sin^2 x$ (where $\sin x \neq 0$):

$\cos x = \sin x$

Divide by $\cos x$ (where $\cos x \neq 0$):

$\tan x = 1$

The general solution for $\tan x = 1$ is $x = n\pi + \frac{\pi}{4}$. For $x \in [0, 2\pi]$, we consider the primary branch:

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

Step 4: Verification of Domain Constraints

We must ensure the arguments of the inverse functions are valid:

• For $\tan^{-1}(\cos x)$: $\cos(\pi/4) = \frac{1}{\sqrt{2}} < 1$. [Condition satisfied]
• For $\tan^{-1}(2\csc x)$: $\csc(\pi/4) = \sqrt{2}$. The argument is $2\sqrt{2}$. Since $\tan^{-1}$ is defined for all real numbers, this is valid.

Geometric Representation (Unit Circle)

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