Given: The statement "$\cot A$ is the product of $\cot$ and $A$".

To Find: Determine whether the given statement is True or False and provide a mathematical justification.

Step 1: Understanding Trigonometric Notation
In trigonometry, the notation $\cot A$ is a shorthand representation for the "cotangent of the angle $A$". Here, $\cot$ is not a separate algebraic variable or a numerical constant, but rather a functional operator (a trigonometric ratio) that acts upon the argument $A$.

Step 2: Analyzing the Relationship
Let us consider a right-angled triangle $\triangle ABC$ where $\angle B = 90^\circ$ and $\angle A$ is one of the acute angles. By definition, the cotangent of angle $A$ is the ratio of the length of the side adjacent to angle $A$ to the length of the side opposite to angle $A$.
Mathematically, $\cot A = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{AB}{BC}$.

Step 3: Evaluating the "Product" Claim
If $\cot A$ were the product of $\cot$ and $A$, then $\cot$ would have to be a value that could be multiplied by $A$. However, $\cot$ by itself has no independent numerical value. It is an operator that requires an angle (the argument) to produce a ratio. If we were to separate them, the expression $\cot$ would be meaningless in the context of geometry and trigonometry.

Step 4: Logical Conclusion
Since $\cot A$ represents a single functional entity where $A$ is the angle associated with the cotangent ratio, it cannot be interpreted as the algebraic product of two distinct factors, $\cot$ and $A$. Therefore, the assertion that $\cot A$ is the product of $\cot$ and $A$ is mathematically incorrect.

Final Answer: False. The term $\cot A$ is a single trigonometric ratio representing the cotangent of angle $A$. The symbol $\cot$ is not a separate variable, and thus $\cot A$ is not the product of $\cot$ and $A$.

Given:

1. $\tan(A + B) = \sqrt{3}$

2. $\tan(A - B) = \frac{1}{\sqrt{3}}$

3. $0^\circ < A + B \le 90^\circ$

4. $A > B$

To find:

The values of angles $A$ and $B$.

Step 1: Determine the angles using trigonometric values.

We know from the standard trigonometric table that $\tan(60^\circ) = \sqrt{3}$.

Given $\tan(A + B) = \sqrt{3}$, we can equate the angles:

$A + B = 60^\circ$ --- (Equation 1)

Similarly, we know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.

Given $\tan(A - B) = \frac{1}{\sqrt{3}}$, we can equate the angles:

$A - B = 30^\circ$ --- (Equation 2)

Step 2: Solve the system of linear equations.

We have a system of two linear equations in two variables:

(1) $A + B = 60^\circ$

(2) $A - B = 30^\circ$

To solve for $A$, we add Equation 1 and Equation 2:

$(A + B) + (A - B) = 60^\circ + 30^\circ$

$A + A + B - B = 90^\circ$

$2A = 90^\circ$

$A = \frac{90^\circ}{2}$

$A = 45^\circ$

Step 3: Substitute the value of $A$ to find $B$.

Substitute $A = 45^\circ$ into Equation 1:

$45^\circ + B = 60^\circ$

$B = 60^\circ - 45^\circ$

$B = 15^\circ$

Step 4: Verification of constraints.

Check if $A > B$: $45^\circ > 15^\circ$ (True).

Check if $0^\circ < A + B \le 90^\circ$: $45^\circ + 15^\circ = 60^\circ$, and $0^\circ < 60^\circ \le 90^\circ$ (True).

Final Answer: A = 45^\circ, B = 15^\circ

Given: $15 \cot A = 8$

To find: $\sin A$ and $\sec A$

Visual Representation:

Step 1: Determine the ratio of sides from the given equation.

Given $15 \cot A = 8$. Dividing both sides by $15$, we get:

$\cot A = \frac{8}{15}$

[Since $\cot A = \frac{\text{Adjacent side}}{\text{Opposite side}}$ in a right-angled triangle]

Let the adjacent side $AB = 8k$ and the opposite side $BC = 15k$, where $k$ is a positive constant.

Step 2: Calculate the hypotenuse using the Pythagoras Theorem.

In $\triangle ABC$, right-angled at $B$:

$AC^2 = AB^2 + BC^2$ [Pythagoras Theorem: Hypotenuse$^2$ = Base$^2$ + Perpendicular$^2$]

$AC^2 = (8k)^2 + (15k)^2$

$AC^2 = 64k^2 + 225k^2$

$AC^2 = 289k^2$

$AC = \sqrt{289k^2} = 17k$

Step 3: Find $\sin A$.

$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC}$

$\sin A = \frac{15k}{17k}$

$\sin A = \frac{15}{17}$

Step 4: Find $\sec A$.

$\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent side}} = \frac{AC}{AB}$

$\sec A = \frac{17k}{8k}$

$\sec A = \frac{17}{8}$

Final Answer: $\sin A = \frac{15}{17}$ and $\sec A = \frac{17}{8}$

Given: In $\triangle ABC$, $\angle B = 90^\circ$ and $\tan A = \frac{1}{\sqrt{3}}$.

To find: The value of $\sin A \cos C + \cos A \sin C$.

Step 1: Determine the sides of the triangle.
In a right-angled triangle, $\tan A = \frac{\text{Opposite side to } A}{\text{Adjacent side to } A} = \frac{BC}{AB}$.
Given $\tan A = \frac{1}{\sqrt{3}}$, we can assume $BC = 1k$ and $AB = \sqrt{3}k$, where $k$ is a positive constant.

Step 2: Calculate the hypotenuse using the Pythagoras Theorem.
The Pythagoras Theorem states: $AC^2 = AB^2 + BC^2$.
$AC^2 = (\sqrt{3}k)^2 + (1k)^2$
$AC^2 = 3k^2 + 1k^2 = 4k^2$
$AC = \sqrt{4k^2} = 2k$.

Step 3: Determine the trigonometric ratios.
For angle $A$:
$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{1k}{2k} = \frac{1}{2}$
$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$

For angle $C$:
$\sin C = \frac{\text{Opposite to } C}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$
$\cos C = \frac{\text{Adjacent to } C}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{1k}{2k} = \frac{1}{2}$

Step 4: Evaluate the expression $\sin A \cos C + \cos A \sin C$.
Substitute the values obtained in Step 3:
$= (\frac{1}{2}) \times (\frac{1}{2}) + (\frac{\sqrt{3}}{2}) \times (\frac{\sqrt{3}}{2})$
$= \frac{1}{4} + \frac{3}{4}$
$= \frac{1 + 3}{4}$
$= \frac{4}{4} = 1$.

Final Answer: 1

Given: In a right-angled triangle $ABC$, where $\angle B = 90^\circ$, we are given that $\sin A = \frac{3}{4}$.

To find: The values of $\cos A$ and $\tan A$.

Step 1: Understanding the Trigonometric Ratio
By definition, in a right-angled triangle, the sine of an angle is the ratio of the side opposite to the angle to the hypotenuse. $\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC}$.
Given $\sin A = \frac{3}{4}$, we can assume $BC = 3k$ and $AC = 4k$, where $k$ is a positive constant.

Step 2: Applying the Pythagoras Theorem
In $\triangle ABC$, by the Pythagoras Theorem: $AC^2 = AB^2 + BC^2$
Substituting the known values: $(4k)^2 = AB^2 + (3k)^2$
$16k^2 = AB^2 + 9k^2$
$AB^2 = 16k^2 - 9k^2$
$AB^2 = 7k^2$
$AB = \sqrt{7k^2} = k\sqrt{7}$

Step 3: Calculating $\cos A$
By definition, $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC}$.
Substituting the values: $\cos A = \frac{k\sqrt{7}}{4k}$
$\cos A = \frac{\sqrt{7}}{4}$

Step 4: Calculating $\tan A$
By definition, $\tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB}$.
Substituting the values: $\tan A = \frac{3k}{k\sqrt{7}}$
$\tan A = \frac{3}{\sqrt{7}}$
To rationalize the denominator: $\tan A = \frac{3}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$

Final Answer: $\cos A = \frac{\sqrt{7}}{4}$ and $\tan A = \frac{3\sqrt{7}}{7}$

Given: The trigonometric statement $\sin(A + B) = \sin A + \sin B$.

To Find: Determine whether the given statement is True or False and provide a justification.

Visual Representation:

Step 1: Understanding the nature of the trigonometric function
The expression $\sin(A + B)$ represents the sine of the sum of two angles $A$ and $B$. In trigonometry, the sine function is a non-linear operator. The distributive property $f(x+y) = f(x) + f(y)$ does not apply to trigonometric functions.

Step 2: Testing the statement with specific values
To verify if the statement is true for all values of $A$ and $B$, we can choose standard angles from the trigonometric table, such as $A = 30^\circ$ and $B = 60^\circ$.

Step 3: Calculating the Left-Hand Side (LHS)
LHS = $\sin(A + B)$
Substitute $A = 30^\circ$ and $B = 60^\circ$:
LHS = $\sin(30^\circ + 60^\circ)$
LHS = $\sin(90^\circ)$
[Since $\sin(90^\circ) = 1$ from the standard trigonometric ratio table]
LHS = $1$

Step 4: Calculating the Right-Hand Side (RHS)
RHS = $\sin A + \sin B$
Substitute $A = 30^\circ$ and $B = 60^\circ$:
RHS = $\sin(30^\circ) + \sin(60^\circ)$
[Using the values $\sin(30^\circ) = \frac{1}{2}$ and $\sin(60^\circ) = \frac{\sqrt{3}}{2}$]
RHS = $\frac{1}{2} + \frac{\sqrt{3}}{2}$
RHS = $\frac{1 + \sqrt{3}}{2}$

Step 5: Comparing LHS and RHS
We observe that:
$1 \neq \frac{1 + \sqrt{3}}{2}$
[Since $\sqrt{3} \approx 1.732$, then $\frac{1 + 1.732}{2} = \frac{2.732}{2} = 1.366$]
Since $1 \neq 1.366$, the LHS is not equal to the RHS.

Conclusion:
Because the equality does not hold for the chosen values of $A$ and $B$, the general statement $\sin(A + B) = \sin A + \sin B$ is mathematically incorrect.

Final Answer: False. The statement is incorrect because the sine function does not distribute over addition. As demonstrated with $A=30^\circ$ and $B=60^\circ$, $\sin(30^\circ+60^\circ) = 1$, whereas $\sin 30^\circ + \sin 60^\circ = \frac{1+\sqrt{3}}{2}$.

Given: The trigonometric function $\cot A$ and the specific angle $A = 0^\circ$.

To Prove/Verify: Whether the statement "$\cot A$ is not defined for $A = 0^\circ$" is True or False.

Step 1: Definition of the Cotangent Function
By the fundamental definitions of trigonometric ratios in a right-angled triangle, the cotangent of an angle $A$ is defined as the reciprocal of the tangent of angle $A$. Mathematically, this is expressed as:
$\cot A = \frac{1}{\tan A}$
Furthermore, since $\tan A = \frac{\sin A}{\cos A}$, we can express $\cot A$ in terms of sine and cosine:
$\cot A = \frac{\cos A}{\sin A}$ [Using the quotient identity for trigonometric functions]

Step 2: Evaluating the expression at $A = 0^\circ$
To determine the value of $\cot 0^\circ$, we substitute $A = 0^\circ$ into the identity derived in Step 1:
$\cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ}$

Step 3: Substituting known trigonometric values
From the standard trigonometric table for specific angles:
$\cos 0^\circ = 1$
$\sin 0^\circ = 0$
Substituting these values into our expression:
$\cot 0^\circ = \frac{1}{0}$

Step 4: Logical Deduction regarding Division by Zero
In the field of real numbers and standard arithmetic, division by zero is undefined. Since the denominator of the fraction $\frac{1}{0}$ is zero, the expression does not yield a finite real number value.
[By the definition of division: $\frac{a}{b} = c \implies a = b \times c$. If $b=0$ and $a \neq 0$, there is no real number $c$ that satisfies the equation $a = 0 \times c$.]

Step 5: Conclusion
Since $\cot 0^\circ$ results in a division by zero, the value is indeed undefined.

Final Answer: True. The statement is true because $\cot A = \frac{\cos A}{\sin A}$, and since $\sin 0^\circ = 0$, the expression $\cot 0^\circ = \frac{1}{0}$ is undefined.

Given: A trigonometric statement $\sin \theta = \frac{4}{3}$ for some angle $\theta$.

To Find: Determine whether the statement is true or false and provide a rigorous justification.

Step 1: Definition of the Sine Function
In a right-angled triangle, for an acute angle $\theta$, the sine function is defined as the ratio of the length of the side opposite to the angle $\theta$ to the length of the hypotenuse.
$\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}}$

Step 2: Properties of a Right-Angled Triangle
In any right-angled triangle, the hypotenuse is the longest side. Let $a$ be the length of the opposite side and $c$ be the length of the hypotenuse. By the geometric property of triangles:
$c > a$ (The hypotenuse must be strictly greater than any other side of the triangle).

Step 3: Analyzing the given ratio
Given $\sin \theta = \frac{4}{3}$.
Comparing this to the definition $\sin \theta = \frac{a}{c}$, we have:
$a = 4$ units
$c = 3$ units

Step 4: Evaluating the validity
According to the property established in Step 2, the hypotenuse ($c$) must be greater than the opposite side ($a$).
Here, $c = 3$ and $a = 4$.
Since $3 < 4$, it implies that $c < a$.
This contradicts the fundamental property of a right-angled triangle where the hypotenuse is the longest side.

Step 5: Theoretical Justification
The range of the sine function for any real angle $\theta$ is restricted to the interval $[-1, 1]$.
Mathematically, $-1 \leq \sin \theta \leq 1$.
Since $\frac{4}{3} \approx 1.33$, and $1.33 > 1$, the value $\frac{4}{3}$ lies outside the possible range of the sine function.

Final Answer: False. The value of $\sin \theta$ cannot exceed 1 because the hypotenuse is always the longest side in a right-angled triangle, making the ratio $\frac{\text{Opposite}}{\text{Hypotenuse}}$ always less than or equal to 1.

Given: A statement $\sin \theta = \cos \theta$ for all values of $\theta$, where $\theta$ is an angle in a right-angled triangle.

To Find/Prove: Determine whether the given statement is True or False and provide a justification.

Step 1: Analyzing the definitions of Sine and Cosine

In a right-angled triangle $ABC$ (right-angled at $B$), for an acute angle $\theta$ at vertex $A$:

$\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC}$

$\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC}$

Step 2: Testing the equality for specific values of $\theta$

The statement claims $\sin \theta = \cos \theta$ for all values of $\theta$. To disprove this, we only need to find one counter-example.

Let $\theta = 0^\circ$:

$\sin 0^\circ = 0$ [From trigonometric table values]

$\cos 0^\circ = 1$ [From trigonometric table values]

Since $0 \neq 1$, the statement $\sin \theta = \cos \theta$ is false for $\theta = 0^\circ$.

Let $\theta = 30^\circ$:

$\sin 30^\circ = \frac{1}{2} = 0.5$

$\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$

Since $0.5 \neq 0.866$, the statement is false for $\theta = 30^\circ$.

Step 3: Identifying the condition where the statement holds

The equation $\sin \theta = \cos \theta$ is only true when $\frac{\sin \theta}{\cos \theta} = 1$, which implies $\tan \theta = 1$.

We know that $\tan 45^\circ = 1$. Therefore, $\sin \theta = \cos \theta$ only when $\theta = 45^\circ$ (within the range $0^\circ \le \theta \le 90^\circ$).

Conclusion:

Since the equality does not hold for all values of $\theta$ (e.g., it fails at $\theta = 0^\circ$ and $\theta = 30^\circ$), the statement is False.

Final Answer: False. The statement $\sin \theta = \cos \theta$ is only true when $\theta = 45^\circ$, not for all values of $\theta$.

Given: The trigonometric ratio $\cot \theta = \frac{7}{8}$.

To Find: The value of $\cot^2 \theta$.

Visual Representation:

Step 1: Understanding the definition of $\cot \theta$

In a right-angled triangle, for an angle $\theta$, the cotangent ratio is defined as the ratio of the length of the adjacent side to the length of the opposite side:

$\cot \theta = \frac{\text{Adjacent side}}{\text{Opposite side}}$

Step 2: Formulating the expression for $\cot^2 \theta$

The expression $\cot^2 \theta$ is mathematically equivalent to $(\cot \theta)^2$. This notation indicates that the entire value of the cotangent of angle $\theta$ must be raised to the power of 2.

Step 3: Substitution and Calculation

Given that $\cot \theta = \frac{7}{8}$, we substitute this value into the expression:

$\cot^2 \theta = (\cot \theta)^2$

$\cot^2 \theta = \left( \frac{7}{8} \right)^2$

Applying the exponent rule $\left( \frac{a}{b} \right)^n = \frac{a^n}{b^n}$:

$\cot^2 \theta = \frac{7^2}{8^2}$

Calculating the squares of the numerator and the denominator:

$7^2 = 7 \times 7 = 49$

$8^2 = 8 \times 8 = 64$

Therefore:

$\cot^2 \theta = \frac{49}{64}$

Final Answer:

Final Answer: \frac{49}{64}

Given: The trigonometric expression $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$.

To find: The numerical value of the given expression.

Visual Representation:

Step 1: Identify the values of the trigonometric ratios.

Based on the standard trigonometric table for specific angles, we have the following values:

• $\sin 60^\circ = \frac{\sqrt{3}}{2}$
• $\cos 30^\circ = \frac{\sqrt{3}}{2}$
• $\sin 30^\circ = \frac{1}{2}$
• $\cos 60^\circ = \frac{1}{2}$

Step 2: Substitute the values into the expression.

The given expression is $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$.

Substituting the values identified in Step 1:

$= \left( \frac{\sqrt{3}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{2} \right) \left( \frac{1}{2} \right)$

Step 3: Perform the multiplication.

Multiply the fractions [Using the rule $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$]:

$= \left( \frac{\sqrt{3} \times \sqrt{3}}{2 \times 2} \right) + \left( \frac{1 \times 1}{2 \times 2} \right)$

$= \left( \frac{3}{4} \right) + \left( \frac{1}{4} \right)$

Step 4: Perform the addition.

Since the denominators are the same, add the numerators [Using the rule $\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}$]:

$= \frac{3 + 1}{4}$

$= \frac{4}{4}$

$= 1$

Final Answer: 1

Given: $3 \cot A = 4$

To Check: Whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$

Step 1: Determine the sides of the right-angled triangle.

Given $3 \cot A = 4$, we can write $\cot A = \frac{4}{3}$.

In a right-angled triangle, $\cot A = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{AB}{BC}$.

Let $AB = 4k$ and $BC = 3k$, where $k$ is a positive constant.

Using the Pythagoras Theorem ($AC^2 = AB^2 + BC^2$):

$AC^2 = (4k)^2 + (3k)^2$

$AC^2 = 16k^2 + 9k^2 = 25k^2$

$AC = \sqrt{25k^2} = 5k$

Step 2: Calculate the trigonometric ratios.

$\tan A = \frac{1}{\cot A} = \frac{3}{4}$

$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5}$

$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}$

Step 3: Evaluate the Left Hand Side (LHS).

LHS = $\frac{1 - \tan^2 A}{1 + \tan^2 A}$

Substitute $\tan A = \frac{3}{4}$:

LHS = $\frac{1 - (\frac{3}{4})^2}{1 + (\frac{3}{4})^2} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}}$

LHS = $\frac{\frac{16-9}{16}}{\frac{16+9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{25}$

Step 4: Evaluate the Right Hand Side (RHS).

RHS = $\cos^2 A - \sin^2 A$

Substitute $\cos A = \frac{4}{5}$ and $\sin A = \frac{3}{5}$:

RHS = $(\frac{4}{5})^2 - (\frac{3}{5})^2$

RHS = $\frac{16}{25} - \frac{9}{25} = \frac{16 - 9}{25} = \frac{7}{25}$

Step 5: Conclusion.

Since LHS = $\frac{7}{25}$ and RHS = $\frac{7}{25}$, the equation holds true.

Final Answer: Yes, $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ is true.

Given: The trigonometric expression $\frac{1 + \tan^2 A}{1 + \cot^2 A}$.

To Find: The simplified value of the given expression by choosing the correct option among the standard trigonometric identities.

Step 1: Recall Fundamental Trigonometric Identities

To simplify the expression, we utilize the following Pythagorean identities:

1. $1 + \tan^2 A = \sec^2 A$ [Since $\sec^2 A - \tan^2 A = 1$]

2. $1 + \cot^2 A = \csc^2 A$ [Since $\csc^2 A - \cot^2 A = 1$]

Step 2: Substitute Identities into the Expression

Substitute the identities identified in Step 1 into the given expression:

$\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A}$

Step 3: Express in terms of Sine and Cosine

Recall the reciprocal relations for trigonometric functions:

$\sec A = \frac{1}{\cos A} \implies \sec^2 A = \frac{1}{\cos^2 A}$

$\csc A = \frac{1}{\sin A} \implies \csc^2 A = \frac{1}{\sin^2 A}$

Substituting these into the expression from Step 2:

$\frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}}$

Step 4: Perform Algebraic Simplification

When dividing fractions, we multiply by the reciprocal of the denominator:

$\frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} = \frac{\sin^2 A}{\cos^2 A}$

Step 5: Apply the Quotient Identity

Recall the quotient identity for tangent:

$\tan A = \frac{\sin A}{\cos A} \implies \tan^2 A = \frac{\sin^2 A}{\cos^2 A}$

Therefore:

$\frac{\sin^2 A}{\cos^2 A} = \tan^2 A$

Conclusion:

The expression $\frac{1 + \tan^2 A}{1 + \cot^2 A}$ simplifies to $\tan^2 A$.

Final Answer: \tan^2 A

Given: $\sec \theta = \frac{13}{12}$

To find: All other trigonometric ratios, i.e., $\sin \theta, \cos \theta, \tan \theta, \csc \theta,$ and $\cot \theta$.

Step 1: Defining the Trigonometric Ratio
In a right-angled triangle, $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent side}}$.
Given $\sec \theta = \frac{13}{12}$, let the Hypotenuse ($AC$) = $13k$ and the Adjacent side ($AB$) = $12k$, where $k$ is a positive constant.

Step 2: Finding the Third Side using Pythagoras Theorem
According to the Pythagoras Theorem: $AC^2 = AB^2 + BC^2$
Substituting the known values:
$(13k)^2 = (12k)^2 + BC^2$
$169k^2 = 144k^2 + BC^2$
$BC^2 = 169k^2 - 144k^2$
$BC^2 = 25k^2$
$BC = \sqrt{25k^2} = 5k$ (Perpendicular/Opposite side)

Step 3: Calculating the Trigonometric Ratios
Using the definitions of trigonometric ratios based on the sides of the triangle:

1. $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}$

2. $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}$

3. $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}$

4. $\csc \theta = \frac{1}{\sin \theta} = \frac{13}{5}$

5. $\cot \theta = \frac{1}{\tan \theta} = \frac{12}{5}$

Final Answer: The trigonometric ratios are $\sin \theta = \frac{5}{13}, \cos \theta = \frac{12}{13}, \tan \theta = \frac{5}{12}, \csc \theta = \frac{13}{5},$ and $\cot \theta = \frac{12}{5}$.

Given: A trigonometric ratio $\sec A = \frac{12}{5}$ for an angle $A$.

To Find: Determine whether the statement "$\sec A = \frac{12}{5}$ for some value of angle $A$" is true or false, and provide a justification.

Step 1: Definition of the Secant Ratio
In a right-angled triangle, for an acute angle $A$, the trigonometric ratio $\sec A$ is defined as the ratio of the length of the hypotenuse to the length of the side adjacent to angle $A$.
$\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent side}}$

Step 2: Analyzing the given value
Given $\sec A = \frac{12}{5}$.
Comparing this to the definition, we can assume:
Hypotenuse $= 12k$
Adjacent side $= 5k$
where $k$ is a positive constant.

Step 3: Applying the Pythagorean Theorem
In any right-angled triangle, the hypotenuse is the longest side. Let the third side (opposite to angle $A$) be $BC$. According to the Pythagorean theorem:
$(\text{Hypotenuse})^2 = (\text{Adjacent side})^2 + (\text{Opposite side})^2$
$(12k)^2 = (5k)^2 + (BC)^2$
$144k^2 = 25k^2 + (BC)^2$
$(BC)^2 = 144k^2 - 25k^2$
$(BC)^2 = 119k^2$
$BC = \sqrt{119}k \approx 10.9k$

Step 4: Logical Justification
Since $12k > 5k$ and $12k > 10.9k$, the hypotenuse is indeed the longest side of the triangle. In a right-angled triangle, the ratio $\frac{\text{Hypotenuse}}{\text{Adjacent}}$ must always be greater than or equal to $1$ because the hypotenuse is always greater than or equal to any other side. Since $\frac{12}{5} = 2.4$, which is greater than $1$, this value is mathematically possible for an angle $A$.

Final Answer: The statement is True. Since the hypotenuse is the longest side in a right-angled triangle, the ratio $\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent}}$ can take any value greater than or equal to $1$. As $\frac{12}{5} = 2.4 > 1$, it is a valid value for $\sec A$.

Given: A trigonometric function $f(\theta) = \cos \theta$, where $\theta$ is an angle in a right-angled triangle, typically considered in the interval $0^\circ \le \theta \le 90^\circ$.

To Determine: Whether the statement "The value of $\cos \theta$ increases as $\theta$ increases" is true or false, with justification.

Step 1: Definition of Cosine
In a right-angled triangle, the cosine of an angle $\theta$ is defined as the ratio of the length of the side adjacent to the angle to the length of the hypotenuse:
$\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$

Step 2: Evaluating specific values of $\cos \theta$
To test the statement, we evaluate $\cos \theta$ at standard angles within the range $0^\circ$ to $90^\circ$:

Step 3: Analyzing the trend
Comparing the values calculated in Step 2:
As $\theta$ increases from $0^\circ$ to $90^\circ$:
$1 > 0.866 > 0.707 > 0.5 > 0$
We observe that as the angle $\theta$ increases, the value of $\cos \theta$ decreases.

Step 4: Justification
In a right-angled triangle, as the angle $\theta$ increases, the side adjacent to $\theta$ decreases in length while the hypotenuse remains constant. Since $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$, a decreasing numerator with a constant denominator results in a decreasing value for the fraction.

Conclusion:
Since the value of $\cos \theta$ decreases as $\theta$ increases from $0^\circ$ to $90^\circ$, the given statement is false.

Final Answer: False. The value of $\cos \theta$ decreases as $\theta$ increases in the interval $0^\circ \le \theta \le 90^\circ$.