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: 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: An algebraic expression involving trigonometric ratios: $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$.

To Find: The numerical value of the given expression.

Step 1: Identification of Trigonometric Values
To evaluate the expression, we must recall the standard trigonometric ratios for the given angles from the trigonometric table:

• $\tan 45^\circ = 1$
• $\cos 30^\circ = \frac{\sqrt{3}}{2}$
• $\sin 60^\circ = \frac{\sqrt{3}}{2}$

Step 2: Substitution of Values into the Expression
Substitute the identified values into the expression $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$:

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

Step 3: Performing Arithmetic Operations
Now, we simplify each term step-by-step:

First, calculate the squares:

• $(1)^2 = 1$
• $\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{(\sqrt{3})^2}{(2)^2} = \frac{3}{4}$

Substitute these back into the expression:

$= 2(1) + \frac{3}{4} - \frac{3}{4}$

Step 4: Final Simplification
Perform the multiplication and addition/subtraction:

$= 2 + \frac{3}{4} - \frac{3}{4}$

[Since $\frac{3}{4} - \frac{3}{4} = 0$]

$= 2 + 0$

$= 2$

Final Answer: 2

Given: The trigonometric equation $\sin 2A = 2 \sin A$.

To find: The value of $A$ for which the given equation holds true, choosing from the standard options usually provided in this context: (A) $0^\circ$, (B) $30^\circ$, (C) $45^\circ$, (D) $60^\circ$.

Visual Representation:

Step 1: Testing Option (A) where $A = 0^\circ$

Substitute $A = 0^\circ$ into the Left Hand Side (LHS) of the equation:

LHS $= \sin 2A = \sin(2 \times 0^\circ) = \sin 0^\circ$

[Since the value of $\sin 0^\circ = 0$ from trigonometric ratio tables]

LHS $= 0$

Now, substitute $A = 0^\circ$ into the Right Hand Side (RHS) of the equation:

RHS $= 2 \sin A = 2 \sin 0^\circ$

[Since $\sin 0^\circ = 0$]

RHS $= 2 \times 0 = 0$

Since LHS = RHS, the equation is true for $A = 0^\circ$.

Step 2: Testing Option (B) where $A = 30^\circ$

LHS $= \sin 2(30^\circ) = \sin 60^\circ$

[Using the standard value $\sin 60^\circ = \frac{\sqrt{3}}{2}$]

LHS $= \frac{\sqrt{3}}{2}$

RHS $= 2 \sin 30^\circ$

[Using the standard value $\sin 30^\circ = \frac{1}{2}$]

RHS $= 2 \times \frac{1}{2} = 1$

Since $\frac{\sqrt{3}}{2} \neq 1$, the equation is false for $A = 30^\circ$.

Step 3: Testing Option (C) where $A = 45^\circ$

LHS $= \sin 2(45^\circ) = \sin 90^\circ$

[Since $\sin 90^\circ = 1$]

LHS $= 1$

RHS $= 2 \sin 45^\circ = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$

Since $1 \neq \sqrt{2}$, the equation is false for $A = 45^\circ$.

Step 4: Testing Option (D) where $A = 60^\circ$

LHS $= \sin 2(60^\circ) = \sin 120^\circ$

[Using the identity $\sin(180^\circ - \theta) = \sin \theta$, $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$]

LHS $= \frac{\sqrt{3}}{2}$

RHS $= 2 \sin 60^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$

Since $\frac{\sqrt{3}}{2} \neq \sqrt{3}$, the equation is false for $A = 60^\circ$.

Conclusion: Comparing the results, the equation $\sin 2A = 2 \sin A$ holds true only when $A = 0^\circ$.

Final Answer: The correct option is (A) $0^\circ$.

Given: The trigonometric expression $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$.

To Find: The numerical value of the given expression.

Visual Representation (Trigonometric Ratios):

Step 1: Identify the values of the trigonometric ratios.
Using the standard trigonometric table for specific angles:

• $\cos 45^\circ = \frac{1}{\sqrt{2}}$
• $\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$
• $\text{cosec } 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{1/2} = 2$

Step 2: Substitute the values into the expression.
The expression is $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$. Substituting the values:

$\text{Expression} = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$

Step 3: Simplify the denominator.
To add the terms in the denominator, find a common denominator:

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

Step 4: Perform the division of fractions.
$\text{Expression} = \frac{1}{\sqrt{2}} \div \frac{2(1 + \sqrt{3})}{\sqrt{3}}$
$\text{Expression} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1 + \sqrt{3})}$
$\text{Expression} = \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} = \frac{\sqrt{3}}{2(\sqrt{2} + \sqrt{6})}$

Step 5: Rationalize the denominator.
Multiply the numerator and denominator by the conjugate $(\sqrt{6} - \sqrt{2})$:

$\text{Expression} = \frac{\sqrt{3}}{2(\sqrt{6} + \sqrt{2})} \times \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}$
$= \frac{\sqrt{18} - \sqrt{6}}{2((\sqrt{6})^2 - (\sqrt{2})^2)}$
$= \frac{3\sqrt{2} - \sqrt{6}}{2(6 - 2)}$
$= \frac{3\sqrt{2} - \sqrt{6}}{2(4)}$
$= \frac{3\sqrt{2} - \sqrt{6}}{8}$

Final Answer: $\frac{3\sqrt{2} - \sqrt{6}}{8}$

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 expression $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$.

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

Step 1: Identify the value of the trigonometric ratio.
From the standard trigonometric table for specific angles:
$\tan 30^\circ = \frac{1}{\sqrt{3}}$

Step 2: Substitute the value into the given expression.
Let the expression be $E$.
$E = \frac{2 \left( \frac{1}{\sqrt{3}} \right)}{1 + \left( \frac{1}{\sqrt{3}} \right)^2}$

Step 3: Simplify the numerator and the denominator.
Numerator: $2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$
Denominator: $1 + \left( \frac{1}{\sqrt{3}} \right)^2 = 1 + \frac{1}{3}$
[Since $(\sqrt{a})^2 = a$]
Denominator: $\frac{3}{3} + \frac{1}{3} = \frac{4}{3}$

Step 4: Perform the division of the fractions.
$E = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$
To divide by a fraction, we multiply by its reciprocal:
$E = \frac{2}{\sqrt{3}} \times \frac{3}{4}$
$E = \frac{2 \times 3}{4 \times \sqrt{3}}$
$E = \frac{6}{4\sqrt{3}}$

Step 5: Simplify the resulting fraction.
$E = \frac{3}{2\sqrt{3}}$
Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$:
$E = \frac{3 \times \sqrt{3}}{2 \times \sqrt{3} \times \sqrt{3}}$
$E = \frac{3\sqrt{3}}{2 \times 3}$
$E = \frac{3\sqrt{3}}{6}$
$E = \frac{\sqrt{3}}{2}$

Step 6: Compare with standard trigonometric values.
We know that:
$\sin 60^\circ = \frac{\sqrt{3}}{2}$
$\cos 60^\circ = \frac{1}{2}$
$\tan 60^\circ = \sqrt{3}$
$\sin 30^\circ = \frac{1}{2}$

Since the calculated value is $\frac{\sqrt{3}}{2}$, it corresponds to $\sin 60^\circ$.

Final Answer: The value of the expression is $\frac{\sqrt{3}}{2}$, which corresponds to $\sin 60^\circ$.

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 $0^\circ \le \theta \le 90^\circ$.

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 $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ}$.

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

Step 1: Recall the standard trigonometric ratio for $30^\circ$.

From the trigonometric table for standard angles, we know that:

$\tan 30^\circ = \frac{1}{\sqrt{3}}$

Step 2: Substitute the value into the given expression.

Let the expression be $E$.

$E = \frac{2 \left( \frac{1}{\sqrt{3}} \right)}{1 - \left( \frac{1}{\sqrt{3}} \right)^2}$

Step 3: Simplify the numerator and the denominator.

Numerator: $2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$

Denominator: $1 - \left( \frac{1}{\sqrt{3}} \right)^2 = 1 - \frac{1}{3}$

[Since $(\sqrt{3})^2 = 3$]

Denominator: $\frac{3 - 1}{3} = \frac{2}{3}$

Step 4: Perform the division of the fractions.

$E = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$

$E = \frac{2}{\sqrt{3}} \times \frac{3}{2}$

[Multiplying by the reciprocal of the denominator]

$E = \frac{2 \times 3}{2 \times \sqrt{3}}$

$E = \frac{3}{\sqrt{3}}$

Step 5: Rationalize the denominator.

$E = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$E = \frac{3\sqrt{3}}{3}$

$E = \sqrt{3}$

Step 6: Compare the result with standard trigonometric values.

We know that $\tan 60^\circ = \sqrt{3}$.

Therefore, the value of the expression is equivalent to $\tan 60^\circ$.

Final Answer: The value of the expression is $\sqrt{3}$, which corresponds to $\tan 60^\circ$.

Given: The trigonometric expression $\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$.

To Find: The numerical value of the given expression.

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

Step 2: Substitute the values into the expression.
Substituting the values identified in Step 1 into the given expression:
$\text{Expression} = \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1}$

Step 3: Simplify the numerator and the denominator.
For the numerator: $\frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}}$
Finding a common denominator ($2\sqrt{3}$):
$\frac{3\sqrt{3} - 4}{2\sqrt{3}}$

For the denominator: $\frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{3}{2}$
Finding a common denominator ($2\sqrt{3}$):
$\frac{4 + 3\sqrt{3}}{2\sqrt{3}}$

Step 4: Perform the division of the fractions.
$\text{Expression} = \frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{3\sqrt{3} + 4}{2\sqrt{3}}}$
Since the denominators are identical, they cancel out:
$\text{Expression} = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$

Step 5: Rationalize the denominator.
To rationalize, multiply the numerator and denominator by the conjugate of the denominator, which is $(3\sqrt{3} - 4)$:
$\frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \times \frac{3\sqrt{3} - 4}{3\sqrt{3} - 4} = \frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3})^2 - (4)^2}$

Applying the algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$ in the numerator and $a^2 - b^2$ in the denominator:
Numerator: $(3\sqrt{3})^2 - 2(3\sqrt{3})(4) + (4)^2 = (9 \times 3) - 24\sqrt{3} + 16 = 27 - 24\sqrt{3} + 16 = 43 - 24\sqrt{3}$
Denominator: $(3\sqrt{3})^2 - (4)^2 = 27 - 16 = 11$

Thus, the expression simplifies to:
$\frac{43 - 24\sqrt{3}}{11}$

Final Answer: \frac{43 - 24\sqrt{3}}{11}

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 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: A trigonometric function $f(\theta) = \sin \theta$, where $\theta$ represents an angle in a right-angled triangle.

To Determine: Whether the statement "The value of $\sin \theta$ increases as $\theta$ increases" is True or False, and provide a justification.

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

Step 2: Evaluating $\sin \theta$ at specific angles in the interval $0^\circ \le \theta \le 90^\circ$
We use the standard trigonometric values for common angles:

Step 3: Analyzing the Trend
By observing the values in the table above:
- At $\theta = 0^\circ$, $\sin \theta = 0$
- At $\theta = 30^\circ$, $\sin \theta = 0.5$
- At $\theta = 45^\circ$, $\sin \theta \approx 0.707$
- At $\theta = 60^\circ$, $\sin \theta \approx 0.866$
- At $\theta = 90^\circ$, $\sin \theta = 1$
As $\theta$ increases from $0^\circ$ to $90^\circ$, the value of $\sin \theta$ consistently increases from $0$ to $1$.

Step 4: Geometric Justification
As the angle $\theta$ increases in a right-angled triangle (while keeping the hypotenuse constant), the length of the side opposite to $\theta$ increases. Since $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$, an increase in the numerator (Opposite side) while the denominator (Hypotenuse) remains fixed results in an increase in the overall value of the fraction.

Conclusion:
Since the value of $\sin \theta$ rises as $\theta$ increases from $0^\circ$ to $90^\circ$, the statement is mathematically correct.

Final Answer: True. The value of $\sin \theta$ increases as $\theta$ increases from $0^\circ$ to $90^\circ$.