Given: The trigonometric identity $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$, where $\theta$ is an acute angle.

To Prove: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$

Step 1: Expressing the identity in terms of $\sin \theta$ and $\cos \theta$

We know the fundamental trigonometric identities: $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$. Substituting these into the Left Hand Side (LHS):

LHS = $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$

Step 2: Simplifying the denominators

Find a common denominator for the terms in the denominators of the fractions:

LHS = $\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$

Step 3: Performing division of fractions

Using the rule $\frac{a/b}{c/d} = \frac{a}{b} \times \frac{d}{c}$:

LHS = $\left( \frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta - \cos \theta} \right) + \left( \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta - \sin \theta} \right)$

LHS = $\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$

Step 4: Aligning the denominators

To combine the fractions, we need a common denominator. Note that $(\cos \theta - \sin \theta) = -(\sin \theta - \cos \theta)$.

LHS = $\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$

Step 5: Combining the fractions

The common denominator is $\sin \theta \cos \theta (\sin \theta - \cos \theta)$:

LHS = $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

Step 6: Applying the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Here, $a = \sin \theta$ and $b = \cos \theta$:

LHS = $\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$

Canceling the common term $(\sin \theta - \cos \theta)$:

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

Step 7: Final simplification

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

LHS = $\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$

LHS = $\frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$

LHS = $\left( \frac{1}{\cos \theta} \right) \left( \frac{1}{\sin \theta} \right) + 1$

LHS = $\sec \theta \text{cosec } \theta + 1$

Conclusion: Since LHS = RHS, the identity is proven.

Final Answer: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$

Given: A right-angled triangle $ABC$ where $\angle B = 90^\circ$, side $AB = 24$ cm, and side $BC = 7$ cm.

To find: The values of $\sin C$ and $\cos C$.

Step 1: Calculating the Hypotenuse ($AC$)

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides [Pythagoras Theorem].

$AC^2 = AB^2 + BC^2$

Substitute the given values:

$AC^2 = (24)^2 + (7)^2$

$AC^2 = 576 + 49$

$AC^2 = 625$

$AC = \sqrt{625} = 25$ cm

Step 2: Defining Trigonometric Ratios for $\angle C$

For $\angle C$, the side opposite is $AB$ and the side adjacent is $BC$. The hypotenuse is $AC$.

$\sin C = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{AB}{AC}$

$\cos C = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AC}$

Step 3: Substituting the values

Using $AB = 24$ cm, $BC = 7$ cm, and $AC = 25$ cm:

$\sin C = \frac{24}{25}$

$\cos C = \frac{7}{25}$

Final Answer: $\sin C = \frac{24}{25}$ and $\cos C = \frac{7}{25}$

Given: An expression $(\sec A + \tan A)(1 - \sin A)$.

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

Step 1: Expressing trigonometric ratios in terms of sine and cosine.

We know the fundamental definitions of trigonometric ratios:

$\sec A = \frac{1}{\cos A}$ [Since secant is the reciprocal of cosine]

$\tan A = \frac{\sin A}{\cos A}$ [Since tangent is the ratio of sine to cosine]

Step 2: Substituting these values into the given expression.

Let the expression be $E$.

$E = \left( \frac{1}{\cos A} + \frac{\sin A}{\cos A} \right) (1 - \sin A)$

Step 3: Simplifying the term inside the parentheses.

Since the denominators are the same, we can combine the fractions:

$E = \left( \frac{1 + \sin A}{\cos A} \right) (1 - \sin A)$

Step 4: Performing the multiplication.

Multiply the numerators together:

$E = \frac{(1 + \sin A)(1 - \sin A)}{\cos A}$

Step 5: Applying the algebraic identity $(a + b)(a - b) = a^2 - b^2$.

Here, $a = 1$ and $b = \sin A$.

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

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

Step 6: Applying the Pythagorean identity.

We know the identity: $\sin^2 A + \cos^2 A = 1$.

Rearranging this gives: $1 - \sin^2 A = \cos^2 A$.

Substituting this into our expression:

$E = \frac{\cos^2 A}{\cos A}$

Step 7: Final simplification.

$E = \frac{\cos A \cdot \cos A}{\cos A}$

$E = \cos A$ [By canceling the common factor $\cos A$ in the numerator and denominator]

Final Answer: The simplified value of the expression is $\cos A$.

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 expression $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$.

To Find: The numerical value of the given expression.

Step 1: Identify the values of the trigonometric ratios.
Based on the standard trigonometric table for specific angles, we have:
$\cos 60^\circ = \frac{1}{2}$
$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$
$\tan 45^\circ = 1$
$\sin 30^\circ = \frac{1}{2}$
$\cos 30^\circ = \frac{\sqrt{3}}{2}$

Step 2: Substitute the values into the numerator.
The numerator is $5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ$.
Substituting the values:
$= 5 \left( \frac{1}{2} \right)^2 + 4 \left( \frac{2}{\sqrt{3}} \right)^2 - (1)^2$
$= 5 \left( \frac{1}{4} \right) + 4 \left( \frac{4}{3} \right) - 1$
$= \frac{5}{4} + \frac{16}{3} - 1$

Step 3: Simplify the numerator.
To add the fractions, find the least common multiple (LCM) of 4 and 3, which is 12.
$= \frac{5 \times 3}{4 \times 3} + \frac{16 \times 4}{3 \times 4} - \frac{1 \times 12}{1 \times 12}$
$= \frac{15}{12} + \frac{64}{12} - \frac{12}{12}$
$= \frac{15 + 64 - 12}{12}$
$= \frac{67}{12}$

Step 4: Simplify the denominator.
The denominator is $\sin^2 30^\circ + \cos^2 30^\circ$.
[Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ for any angle $\theta$]:
$\sin^2 30^\circ + \cos^2 30^\circ = 1$
Alternatively, by calculation:
$= \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2$
$= \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$

Step 5: Calculate the final value.
The expression is $\frac{\text{Numerator}}{\text{Denominator}} = \frac{67/12}{1}$.
$= \frac{67}{12}$

Final Answer: \frac{67}{12}

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: $\cot \theta = \frac{7}{8}$

To find: The value of $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$

Step 1: Define the trigonometric ratio in terms of sides.
In a right-angled triangle, $\cot \theta = \frac{\text{Adjacent side}}{\text{Opposite side}}$.
Given $\cot \theta = \frac{7}{8}$, let the adjacent side $AB = 7k$ and the opposite side $BC = 8k$, where $k$ is a positive constant.

Step 2: Calculate the hypotenuse using the Pythagorean Theorem.
According to the Pythagorean Theorem: $AC^2 = AB^2 + BC^2$.
$AC^2 = (7k)^2 + (8k)^2$
$AC^2 = 49k^2 + 64k^2$
$AC^2 = 113k^2$
$AC = \sqrt{113}k$

Step 3: Determine $\sin \theta$ and $\cos \theta$.
$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}}$
$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}}$

Step 4: Simplify the expression using algebraic identities.
The expression is $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$.
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$:
Numerator: $(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta$
Denominator: $(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta$
Expression = $\frac{1 - \sin^2 \theta}{1 - \cos^2 \theta}$

Step 5: Substitute the values and calculate.
$1 - \sin^2 \theta = 1 - (\frac{8}{\sqrt{113}})^2 = 1 - \frac{64}{113} = \frac{113 - 64}{113} = \frac{49}{113}$
$1 - \cos^2 \theta = 1 - (\frac{7}{\sqrt{113}})^2 = 1 - \frac{49}{113} = \frac{113 - 49}{113} = \frac{64}{113}$
Expression = $\frac{49/113}{64/113} = \frac{49}{113} \times \frac{113}{64} = \frac{49}{64}$

Final Answer: \frac{49}{64}

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: The statement "$\cos A$ is the abbreviation used for the cosecant of angle $A$."

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

Step 1: Defining Trigonometric Abbreviations

In trigonometry, the six primary trigonometric functions are defined by specific abbreviations derived from the names of the ratios of sides in a right-angled triangle. Let us define them formally:

• $\sin A$: Abbreviation for sine of angle $A$.
• $\cos A$: Abbreviation for cosine of angle $A$.
• $\tan A$: Abbreviation for tangent of angle $A$.
• $\cot A$: Abbreviation for cotangent of angle $A$.
• $\sec A$: Abbreviation for secant of angle $A$.
• $\csc A$ (or $\text{cosec } A$): Abbreviation for cosecant of angle $A$.

Step 2: Analyzing the Statement

The statement claims that $\cos A$ is the abbreviation for the cosecant of angle $A$. Based on the definitions established in Step 1:

1. The abbreviation $\cos A$ corresponds to the cosine of angle $A$.

2. The abbreviation for the cosecant of angle $A$ is $\csc A$ or $\text{cosec } A$.

Step 3: Logical Comparison

Since $\cos A$ represents the cosine function and not the cosecant function, the statement provided is mathematically incorrect. The term "cosecant" is the reciprocal of the sine function ($\csc A = \frac{1}{\sin A}$), whereas "cosine" is a distinct trigonometric ratio ($\cos A = \frac{\text{adjacent side}}{\text{hypotenuse}}$).

Conclusion:

The statement is False because $\cos A$ is the abbreviation for the cosine of angle $A$, while the abbreviation for the cosecant of angle $A$ is $\text{cosec } A$.

Final Answer: False. $\cos A$ is the abbreviation for the cosine of angle $A$, whereas the abbreviation for the cosecant of angle $A$ is $\text{cosec } A$.

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: A statement regarding the trigonometric ratio $\tan A$, where $A$ is an acute angle in a right-angled triangle.

To Find/Prove: Determine whether the statement "The value of $\tan A$ is always less than 1" is True or False, and provide a mathematical justification.

Visual Representation:

Step 1: Definition of Tangent Ratio

In a right-angled triangle $ABC$ right-angled at $B$, the tangent of angle $A$ is defined as the ratio of the length of the side opposite to angle $A$ to the length of the side adjacent to angle $A$.

Mathematically: $\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB}$

Step 2: Analyzing the Ratio

In a right-angled triangle, the lengths of the sides $BC$ (opposite) and $AB$ (adjacent) are independent of each other, provided they are both positive real numbers. There is no geometric constraint that forces the opposite side to be smaller than the adjacent side.

Step 3: Counter-Example Verification

Let us consider a specific case where the opposite side is greater than the adjacent side. Suppose $BC = 4$ units and $AB = 3$ units.

Then, $\tan A = \frac{4}{3}$

Performing the division: $\frac{4}{3} = 1.333...$

Since $1.333... > 1$, we have found a case where $\tan A$ is greater than 1.

Step 4: Theoretical Justification

The tangent function $\tan \theta$ is defined for $0^\circ \le \theta < 90^\circ$. As $\theta$ approaches $90^\circ$, the value of $\tan \theta$ increases without bound (tends to infinity). For example, $\tan 60^\circ = \sqrt{3} \approx 1.732$, which is clearly greater than 1.

Conclusion:

Since there exist values of $A$ for which $\tan A > 1$, the statement "The value of $\tan A$ is always less than 1" is incorrect.

Final Answer: False. The value of $\tan A$ can take any real value. For instance, if $\angle A = 60^\circ$, then $\tan 60^\circ = \sqrt{3} \approx 1.732$, which is greater than 1.

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: 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: The trigonometric expression $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}$.

To Find: The numerical value of the expression and identify the correct option among the standard trigonometric values.

Step 1: Recall the trigonometric ratio for $45^\circ$
From the standard trigonometric table for specific angles, we know that:
$\tan 45^\circ = 1$

Step 2: Substitute the value into the expression
The given expression is:
$E = \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}$

Substituting $\tan 45^\circ = 1$ into the expression:
$E = \frac{1 - (1)^2}{1 + (1)^2}$

Step 3: Perform arithmetic simplification
Calculate the square of the value:
$(1)^2 = 1 \times 1 = 1$

Substitute this back into the fraction:
$E = \frac{1 - 1}{1 + 1}$

Perform the subtraction in the numerator and the addition in the denominator:
$E = \frac{0}{2}$

Step 4: Final evaluation
Any fraction with a numerator of $0$ and a non-zero denominator is equal to $0$.
$E = 0$

Step 5: Justification and Comparison with Options
We evaluate the standard trigonometric values typically provided in such multiple-choice questions:
(A) $\tan 90^\circ$ (Undefined)
(B) $1$
(C) $\sin 45^\circ = \frac{1}{\sqrt{2}}$
(D) $0$

Since our calculated value is $0$, the expression is equal to $0$.

Final Answer: The value of the expression is 0.

Given: The trigonometric identity $(\text{cosec } A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$, where $A$ is an acute angle.

To Prove: The Left Hand Side (LHS) is equal to the Right Hand Side (RHS).

Step 1: Simplifying the Left Hand Side (LHS)

LHS = $(\text{cosec } A - \sin A)(\sec A - \cos A)$

Using the reciprocal identities $\text{cosec } A = \frac{1}{\sin A}$ and $\sec A = \frac{1}{\cos A}$:

LHS = $\left( \frac{1}{\sin A} - \sin A \right) \left( \frac{1}{\cos A} - \cos A \right)$

Taking the common denominator for each bracket:

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

Applying the Pythagorean identity $\sin^2 A + \cos^2 A = 1$, which implies $1 - \sin^2 A = \cos^2 A$ and $1 - \cos^2 A = \sin^2 A$:

LHS = $\left( \frac{\cos^2 A}{\sin A} \right) \left( \frac{\sin^2 A}{\cos A} \right)$

Canceling the common terms in the numerator and denominator:

LHS = $\cos A \cdot \sin A$

Step 2: Simplifying the Right Hand Side (RHS)

RHS = $\frac{1}{\tan A + \cot A}$

Using the quotient identities $\tan A = \frac{\sin A}{\cos A}$ and $\cot A = \frac{\cos A}{\sin A}$:

RHS = $\frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$

Finding the common denominator in the denominator of the fraction:

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

Applying the Pythagorean identity $\sin^2 A + \cos^2 A = 1$:

RHS = $\frac{1}{\frac{1}{\cos A \sin A}}$

By the property of reciprocals of fractions ($\frac{1}{1/x} = x$):

RHS = $\sin A \cos A$

Step 3: Conclusion

Since LHS = $\sin A \cos A$ and RHS = $\sin A \cos A$, we have shown that LHS = RHS.

Final Answer: Hence, it is proved that $(\text{cosec } A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$.