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:

In $\triangle PQR$, $\angle Q = 90^\circ$.

The length of side $PQ = 5$ cm.

The sum of the lengths of the hypotenuse and the other side is $PR + QR = 25$ cm.

To Find:

The values of $\sin P$, $\cos P$, and $\tan P$.

Step 1: Expressing sides in terms of a single variable

Let $QR = x$ cm. Since $PR + QR = 25$ cm, we can express $PR$ as:

$PR = 25 - x$

Step 2: Applying the Pythagorean Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides [Pythagorean Theorem: $Hypotenuse^2 = Base^2 + Perpendicular^2$].

$PR^2 = PQ^2 + QR^2$

Substitute the known values and expressions:

$(25 - x)^2 = 5^2 + x^2$

Step 3: Solving for $x$

Expand the left side using the identity $(a - b)^2 = a^2 - 2ab + b^2$:

$625 - 50x + x^2 = 25 + x^2$

Subtract $x^2$ from both sides:

$625 - 50x = 25$

Rearrange to solve for $x$:

$625 - 25 = 50x$

$600 = 50x$

$x = \frac{600}{50} = 12$

Thus, $QR = 12$ cm.

Now, find $PR$: $PR = 25 - 12 = 13$ cm.

Step 4: Calculating Trigonometric Ratios

For $\angle P$, the side opposite is $QR = 12$ cm, the side adjacent is $PQ = 5$ cm, and the hypotenuse is $PR = 13$ cm.

Using the definitions of trigonometric ratios:

$\sin P = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{12}{13}$

$\cos P = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{5}{13}$

$\tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ} = \frac{12}{5}$

Final Answer:

$\sin P = \frac{12}{13}, \cos P = \frac{5}{13}, \tan P = \frac{12}{5}$

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 expression $(1 + \tan \theta + \sec \theta)(1 + \cot \theta - \text{cosec } \theta)$.

To Find: The simplified value of the given expression.

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

We use the following fundamental trigonometric identities:

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\sec \theta = \frac{1}{\cos \theta}$

$\cot \theta = \frac{\cos \theta}{\sin \theta}$

$\text{cosec } \theta = \frac{1}{\sin \theta}$

Substituting these into the expression:

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

Step 2: Simplifying the terms inside each bracket.

For the first bracket, find a common denominator ($\cos \theta$):

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

For the second bracket, find a common denominator ($\sin \theta$):

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

Step 3: Multiplying the two fractions.

Expression $= \frac{(\sin \theta + \cos \theta + 1)(\sin \theta + \cos \theta - 1)}{\sin \theta \cos \theta}$

Step 4: Applying the algebraic identity $(a + b)(a - b) = a^2 - b^2$.

Let $a = (\sin \theta + \cos \theta)$ and $b = 1$.

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

Expanding $(\sin \theta + \cos \theta)^2$ using $(a + b)^2 = a^2 + b^2 + 2ab$:

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

Step 5: Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$.

Numerator $= (1 + 2 \sin \theta \cos \theta) - 1$

Numerator $= 2 \sin \theta \cos \theta$

Step 6: Final simplification.

Expression $= \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}$

Canceling the common terms $\sin \theta \cos \theta$ (assuming $\sin \theta \neq 0$ and $\cos \theta \neq 0$):

Expression $= 2$

Final Answer: 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: 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: An algebraic expression involving trigonometric functions: $9 \sec^2 A - 9 \tan^2 A$.

To Find: The numerical value of the expression by simplifying it using trigonometric identities.

Step 1: Factoring the expression
The given expression is $9 \sec^2 A - 9 \tan^2 A$. We observe that the constant $9$ is common to both terms. We can factor it out using the distributive property of multiplication over subtraction:
$9 \sec^2 A - 9 \tan^2 A = 9(\sec^2 A - \tan^2 A)$

Step 2: Identifying the relevant trigonometric identity
We recall the fundamental trigonometric identity relating the secant and tangent functions:
$1 + \tan^2 A = \sec^2 A$
[Since the identity holds for all values of $A$ where the functions are defined]

Step 3: Rearranging the identity
To match the expression inside the parentheses, we rearrange the identity $1 + \tan^2 A = \sec^2 A$ by subtracting $\tan^2 A$ from both sides:
$1 = \sec^2 A - \tan^2 A$
[By the subtraction property of equality]

Step 4: Substitution and Final Calculation
Now, substitute the value of $(\sec^2 A - \tan^2 A)$ into the factored expression from Step 1:
$9(\sec^2 A - \tan^2 A) = 9(1)$
[Substituting $1$ for the expression $\sec^2 A - \tan^2 A$]
$9(1) = 9$

Justification: The expression simplifies to $9$ because the difference between the square of the secant and the square of the tangent of an angle is always equal to $1$, as derived from the Pythagorean identity $1 + \tan^2 A = \sec^2 A$.

Final Answer: 9

Given: An trigonometric expression involving an acute angle $\theta$: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$.

To Prove: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$.

Step 1: Analyze the Left-Hand Side (LHS)

The given expression is: $LHS = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$.

Step 2: Factor out common terms from the numerator and denominator

In the numerator, $\sin \theta$ is common to both terms. In the denominator, $\cos \theta$ is common to both terms.

Factoring the numerator: $\sin \theta - 2 \sin^3 \theta = \sin \theta (1 - 2 \sin^2 \theta)$.

Factoring the denominator: $2 \cos^3 \theta - \cos \theta = \cos \theta (2 \cos^2 \theta - 1)$.

Substituting these back into the LHS expression:

$LHS = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)}$

Step 3: Apply the Trigonometric Identity

Recall the fundamental Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$.

From this identity, we can express $1$ as $(\sin^2 \theta + \cos^2 \theta)$.

Substitute this into the numerator term $(1 - 2 \sin^2 \theta)$:

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

$= \cos^2 \theta - \sin^2 \theta$

Now, substitute this into the denominator term $(2 \cos^2 \theta - 1)$:

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

$= 2 \cos^2 \theta - \sin^2 \theta - \cos^2 \theta$

$= \cos^2 \theta - \sin^2 \theta$

Step 4: Simplify the expression

Substitute the simplified terms back into the LHS:

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

Since $\theta$ is an acute angle and the expression is defined, $(\cos^2 \theta - \sin^2 \theta) \neq 0$. Therefore, we can cancel the common factor $(\cos^2 \theta - \sin^2 \theta)$ from the numerator and denominator:

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

Step 5: Apply the Quotient Identity

Using the quotient identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$LHS = \tan \theta$

Conclusion:

Since $LHS = RHS$, the identity is proven.

Final Answer: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$ is proved.

Given: An identity involving an acute angle $A$, specifically $\sqrt{\frac{1 + \sin A}{1 – \sin A}}$.

To Prove: $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$

Step 1: Consider the Left Hand Side (LHS) of the identity.

LHS = $\sqrt{\frac{1 + \sin A}{1 - \sin A}}$

Step 2: Rationalize the denominator inside the square root.

To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator, which is $(1 + \sin A)$.

LHS = $\sqrt{\frac{(1 + \sin A) \times (1 + \sin A)}{(1 - \sin A) \times (1 + \sin A)}}$

Step 3: Simplify the expression using algebraic identities.

In the numerator, we have $(1 + \sin A)(1 + \sin A) = (1 + \sin A)^2$.

In the denominator, we use the difference of squares identity: $(a - b)(a + b) = a^2 - b^2$.

Here, $(1 - \sin A)(1 + \sin A) = 1^2 - \sin^2 A = 1 - \sin^2 A$.

LHS = $\sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$

Step 4: Apply the fundamental trigonometric identity.

We know that $\sin^2 A + \cos^2 A = 1$, which implies that $1 - \sin^2 A = \cos^2 A$.

Substituting this into our expression:

LHS = $\sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$

Step 5: Extract the square root.

Since the square root of a squared term is the term itself (given $A$ is an acute angle, $\cos A > 0$ and $1 + \sin A > 0$):

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

Step 6: Separate the terms in the fraction.

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

Step 7: Apply trigonometric reciprocal and quotient identities.

We know that $\frac{1}{\cos A} = \sec A$ and $\frac{\sin A}{\cos A} = \tan A$.

LHS = $\sec A + \tan A$

Conclusion:

Since the Left Hand Side (LHS) is equal to the Right Hand Side (RHS), the identity is proven.

Final Answer: $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$ is proved.

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

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 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: 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 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: 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: 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