Given: The trigonometric ratios $\sin A$, $\sec A$, and $\tan A$.

To Find: Express $\sin A$, $\sec A$, and $\tan A$ in terms of $\cot A$.

Step 1: Expressing $\tan A$ in terms of $\cot A$

By the definition of reciprocal trigonometric identities, we know that the tangent ratio is the reciprocal of the cotangent ratio.

Formula: $\tan A = \frac{1}{\cot A}$

Thus, $\tan A = \frac{1}{\cot A}$.

Step 2: Expressing $\sin A$ in terms of $\cot A$

We use the trigonometric identity relating $\csc A$ and $\cot A$: $1 + \cot^2 A = \csc^2 A$.

Taking the square root on both sides: $\csc A = \sqrt{1 + \cot^2 A}$.

Since $\sin A = \frac{1}{\csc A}$ [Reciprocal identity], we substitute the expression for $\csc A$:

$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$

Step 3: Expressing $\sec A$ in terms of $\cot A$

We use the trigonometric identity relating $\sec A$ and $\tan A$: $1 + \tan^2 A = \sec^2 A$.

Substitute the expression for $\tan A$ found in Step 1 ($\tan A = \frac{1}{\cot A}$):

$\sec^2 A = 1 + \left(\frac{1}{\cot A}\right)^2$

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

Find a common denominator:

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

Taking the square root on both sides:

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

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

Final Answer:

$\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$

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

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

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:

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 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 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 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: The trigonometric expression $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$.

To Prove: $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A$, using the identity $\text{cosec}^2 A = 1 + \cot^2 A$.

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

We begin with the expression:
$LHS = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$

To express the terms in $\text{cosec } A$ and $\cot A$, we divide both the numerator and the denominator by $\sin A$.
$LHS = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}}$

Step 2: Simplifying using Trigonometric Ratios

Recall the definitions: $\cot A = \frac{\cos A}{\sin A}$ and $\text{cosec } A = \frac{1}{\sin A}$.
Substituting these into the expression:
$LHS = \frac{\cot A - 1 + \text{cosec } A}{\cot A + 1 - \text{cosec } A}$

Step 3: Applying the Trigonometric Identity

We are given the identity $\text{cosec}^2 A = 1 + \cot^2 A$. This can be rearranged as:
$1 = \text{cosec}^2 A - \cot^2 A$
We substitute this value of $1$ into the denominator of our expression:
$LHS = \frac{\cot A + \text{cosec } A - 1}{\cot A - \text{cosec } A + (\text{cosec}^2 A - \cot^2 A)}$

Step 4: Factoring the Denominator

Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$, we factor $(\text{cosec}^2 A - \cot^2 A)$:
$LHS = \frac{\cot A + \text{cosec } A - 1}{\cot A - \text{cosec } A + (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}$

To simplify, we factor out $(\cot A + \text{cosec } A)$ from the denominator. Note that $(\cot A - \text{cosec } A) = -(\text{cosec } A - \cot A)$:
$LHS = \frac{\cot A + \text{cosec } A - 1}{-(\text{cosec } A - \cot A) + (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}$

Step 5: Final Cancellation

Factor out $(\text{cosec } A - \cot A)$ in the denominator:
$LHS = \frac{\cot A + \text{cosec } A - 1}{(\text{cosec } A - \cot A) [(\text{cosec } A + \cot A) - 1]}$

Since the term $(\cot A + \text{cosec } A - 1)$ appears in both the numerator and the denominator, they cancel out:
$LHS = \frac{1}{\text{cosec } A - \cot A}$

To reach the final form, multiply the numerator and denominator by the conjugate $(\text{cosec } A + \cot A)$:
$LHS = \frac{1 \cdot (\text{cosec } A + \cot A)}{(\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}$
$LHS = \frac{\text{cosec } A + \cot A}{\text{cosec}^2 A - \cot^2 A}$

Since $\text{cosec}^2 A - \cot^2 A = 1$:
$LHS = \text{cosec } A + \cot A = RHS$

Final Answer: Since the Left-Hand Side equals the Right-Hand Side, the identity $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A$ is proved.

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: An identity involving trigonometric ratios of an acute angle $A$: $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$.

To Prove: 1. $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$ 2. $\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

Step 1: Proving the first part $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$

We use the fundamental trigonometric identities:
$1 + \tan^2 A = \sec^2 A$
$1 + \cot^2 A = \csc^2 A$
[Since $\sec^2 \theta - \tan^2 \theta = 1$ and $\csc^2 \theta - \cot^2 \theta = 1$]

Substitute these into the expression:
$\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A}$

Express $\sec A$ and $\csc A$ in terms of $\sin A$ and $\cos A$:
$\sec A = \frac{1}{\cos A}$ and $\csc A = \frac{1}{\sin A}$

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

Using the identity $\tan A = \frac{\sin A}{\cos A}$:
$\frac{\sin^2 A}{\cos^2 A} = \tan^2 A$

Thus, the first part is proved: $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$.

Step 2: Proving the second part $\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

Express $\cot A$ in terms of $\tan A$:
$\cot A = \frac{1}{\tan A}$

Substitute this into the expression:
$\left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}}\right)^2 = \left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}\right)^2$

Simplify the fraction:
$\left((1 - \tan A) \times \frac{\tan A}{\tan A - 1}\right)^2$

Note that $(1 - \tan A) = -( \tan A - 1)$:
$\left(\frac{-( \tan A - 1) \times \tan A}{\tan A - 1}\right)^2 = (-\tan A)^2$

Calculate the square:
$(-\tan A)^2 = \tan^2 A$

Conclusion:
Since both the first expression and the second expression are equal to $\tan^2 A$, the identity is proved:
$\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

Final Answer: The identity is proven as both sides simplify to $\tan^2 A$.

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 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 $(\text{cosec } A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$.

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: A right-angled triangle $PQR$ where $\angle Q = 90^\circ$. The length of side $PQ = 12\text{ cm}$ and the length of the hypotenuse $PR = 13\text{ cm}$.

To Find: The value of the expression $\tan P - \cot R$.

Step 1: Determine the length of side $QR$ using the Pythagoras Theorem.

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

Substituting the given values:

$13^2 = 12^2 + QR^2$

$169 = 144 + QR^2$

$QR^2 = 169 - 144$

$QR^2 = 25$

$QR = \sqrt{25} = 5\text{ cm}$

Step 2: Calculate $\tan P$.

For $\angle P$, the side opposite is $QR$ and the side adjacent is $PQ$.

The trigonometric ratio for tangent is defined as: $\tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ}$

$\tan P = \frac{5}{12}$

Step 3: Calculate $\cot R$.

For $\angle R$, the side opposite is $PQ$ and the side adjacent is $QR$.

The trigonometric ratio for cotangent is defined as: $\cot R = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{QR}{PQ}$

$\cot R = \frac{5}{12}$

Step 4: Evaluate the expression $\tan P - \cot R$.

Substitute the values obtained in Step 2 and Step 3:

$\tan P - \cot R = \frac{5}{12} - \frac{5}{12}$

$\tan P - \cot R = 0$

Final Answer: 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: 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 $\triangle ABC$, $\angle B = 90^\circ$, $AB = 24\text{ cm}$, and $BC = 7\text{ cm}$.

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

Step 1: Determine the length of the hypotenuse ($AC$).

Since $\triangle ABC$ is a right-angled triangle, we apply the Pythagoras Theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

$AC^2 = AB^2 + BC^2$ [Pythagoras Theorem]

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

$AC^2 = 576 + 49$

$AC^2 = 625$

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

Step 2: Define the trigonometric ratios for angle $A$.

For $\angle A$, the side opposite is $BC = 7\text{ cm}$, the side adjacent is $AB = 24\text{ cm}$, and the hypotenuse is $AC = 25\text{ cm}$.

The definitions of sine and cosine are:

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

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

Step 3: Calculate the values.

Substituting the known lengths into the ratios:

$\sin A = \frac{7}{25}$

$\cos A = \frac{24}{25}$

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