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

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: 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: The trigonometric identity $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$, 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)

The expression is given as: $LHS = \frac{1 + \sec A}{\sec A}$

We know the fundamental trigonometric identity relating secant and cosine: $\sec A = \frac{1}{\cos A}$ [Definition of secant as the reciprocal of cosine].

Substituting this into the LHS expression:

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

To simplify the numerator, find a common denominator:

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

Multiplying by the reciprocal of the denominator:

$LHS = \left( \frac{\cos A + 1}{\cos A} \right) \times \left( \frac{\cos A}{1} \right)$

Canceling the common term $\cos A$ in the numerator and denominator:

$LHS = 1 + \cos A$

Step 2: Simplifying the Right Hand Side (RHS)

The expression is given as: $RHS = \frac{\sin^2 A}{1 – \cos A}$

We use the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$, which implies $\sin^2 A = 1 - \cos^2 A$ [Using the identity $\sin^2 \theta + \cos^2 \theta = 1$].

Substituting this into the RHS expression:

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

Recognizing the numerator as a difference of squares, where $a^2 - b^2 = (a - b)(a + b)$. Here, $1 - \cos^2 A = (1)^2 - (\cos A)^2 = (1 - \cos A)(1 + \cos A)$:

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

Canceling the common term $(1 - \cos A)$ in the numerator and denominator (assuming $1 - \cos A \neq 0$, which is true for acute angle $A \neq 0^\circ$):

$RHS = 1 + \cos A$

Step 3: Conclusion

Since $LHS = 1 + \cos A$ and $RHS = 1 + \cos A$, we have shown that $LHS = RHS$.

Final Answer: Since both sides simplify to $1 + \cos A$, the identity $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$ is proven.

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: 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: An identity involving trigonometric ratios of an acute angle $A$: $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}$.

To Prove: $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$.

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

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

Step 2: Finding a common denominator.

To add the two fractions, we find the common denominator, which is $(1 + \sin A)(\cos A)$.

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

Step 3: Expanding the numerator.

LHS = $\frac{\cos^2 A + (1 + 2\sin A + \sin^2 A)}{(1 + \sin A)(\cos A)}$ [Using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$ where $a=1, b=\sin A$]

Step 4: Applying the Pythagorean trigonometric identity.

We know that $\sin^2 A + \cos^2 A = 1$. Rearranging the terms in the numerator:

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

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

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

Step 5: Simplifying the expression.

Factor out the constant 2 from the numerator:

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

Cancel the common term $(1 + \sin A)$ from the numerator and the denominator:

LHS = $\frac{2}{\cos A}$

Step 6: Converting to the final form.

Using the reciprocal identity $\frac{1}{\cos A} = \sec A$:

LHS = $2 \sec A$

Since LHS = RHS, the identity is proven.

Final Answer: The identity $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$ is successfully proven.

Given: An identity involving trigonometric functions of an acute angle $\theta$, specifically $(\text{cosec } \theta - \cot \theta)^2$.

To Prove: $(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$

Step 1: Expressing the Left Hand Side (LHS) in terms of Sine and Cosine.

The given expression is $LHS = (\text{cosec } \theta - \cot \theta)^2$.

We use the fundamental trigonometric identities:
$\text{cosec } \theta = \frac{1}{\sin \theta}$
$\cot \theta = \frac{\cos \theta}{\sin \theta}$

Substituting these into the LHS:

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

Step 2: Simplifying the fraction inside the square.

Since the denominators are the same, we combine the terms:

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

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

Step 3: Applying the Pythagorean Identity.

We know the identity $\sin^2 \theta + \cos^2 \theta = 1$, which implies $\sin^2 \theta = 1 - \cos^2 \theta$.

Substituting this into the denominator:

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

Step 4: Factoring the denominator.

The denominator is in the form of a difference of squares, $a^2 - b^2 = (a - b)(a + b)$, where $a = 1$ and $b = \cos \theta$.

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

Substituting this back into the expression:

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

Step 5: Final Cancellation.

Canceling the common term $(1 - \cos \theta)$ from the numerator and the denominator:

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

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

Final Answer: Since the Left Hand Side simplifies to $\frac{1 - \cos \theta}{1 + \cos \theta}$, which is equal to the Right Hand Side, the identity $(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$ is proven.

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: An identity involving trigonometric functions of an acute angle $A$: $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2$.

To Prove: $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$.

Step 1: Expanding the Left Hand Side (LHS)

We start with the expression: $LHS = (\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2$.

Using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$, we expand both terms:

$(\sin A + \text{cosec } A)^2 = \sin^2 A + 2\sin A \cdot \text{cosec } A + \text{cosec}^2 A$

$(\cos A + \sec A)^2 = \cos^2 A + 2\cos A \cdot \sec A + \sec^2 A$

Step 2: Simplifying the expanded terms

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

Substituting these into the middle terms:

$2\sin A \cdot \text{cosec } A = 2\sin A \cdot \left(\frac{1}{\sin A}\right) = 2$

$2\cos A \cdot \sec A = 2\cos A \cdot \left(\frac{1}{\cos A}\right) = 2$

Now, substitute these back into the expanded expression:

$LHS = \sin^2 A + 2 + \text{cosec}^2 A + \cos^2 A + 2 + \sec^2 A$

Step 3: Grouping terms using Trigonometric Identities

Rearrange the terms to group $\sin^2 A$ and $\cos^2 A$ together:

$LHS = (\sin^2 A + \cos^2 A) + 2 + 2 + \text{cosec}^2 A + \sec^2 A$

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

$LHS = 1 + 4 + \text{cosec}^2 A + \sec^2 A$

$LHS = 5 + \text{cosec}^2 A + \sec^2 A$

Step 4: Converting to $\tan^2 A$ and $\cot^2 A$

We use the trigonometric identities: $\text{cosec}^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$.

Substitute these into the expression:

$LHS = 5 + (1 + \cot^2 A) + (1 + \tan^2 A)$

$LHS = 5 + 1 + 1 + \tan^2 A + \cot^2 A$

$LHS = 7 + \tan^2 A + \cot^2 A$

Step 5: Conclusion

Since the simplified LHS is equal to the Right Hand Side (RHS):

$LHS = RHS = 7 + \tan^2 A + \cot^2 A$

Final Answer: The identity $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$ is proven.

Given: An angle $A$ in a right-angled triangle, where the trigonometric ratio is expressed in terms of $\sec A$.

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

Step 1: Expressing $\cos A$ in terms of $\sec A$

By the definition of reciprocal trigonometric ratios, we know that $\cos A$ is the reciprocal of $\sec A$.

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

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

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

$\sin^2 A = 1 - \cos^2 A$ [Subtracting $\cos^2 A$ from both sides]

$\sin^2 A = 1 - \left(\frac{1}{\sec A}\right)^2$ [Substituting $\cos A = \frac{1}{\sec A}$]

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

$\sin A = \sqrt{\frac{\sec^2 A - 1}{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$

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

Using the identity: $1 + \tan^2 A = \sec^2 A$.

$\tan^2 A = \sec^2 A - 1$ [Subtracting 1 from both sides]

$\tan A = \sqrt{\sec^2 A - 1}$

Step 4: Expressing $\csc A$ in terms of $\sec A$

By definition, $\csc A = \frac{1}{\sin A}$.

$\csc A = \frac{1}{\frac{\sqrt{\sec^2 A - 1}}{\sec A}}$ [Substituting the expression for $\sin A$ derived in Step 2]

$\csc A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$

Step 5: Expressing $\cot A$ in terms of $\sec A$

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

$\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$ [Substituting the expression for $\tan A$ derived in Step 3]

Final Answer:

The trigonometric ratios in terms of $\sec A$ are:

$\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$

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

$\tan A = \sqrt{\sec^2 A - 1}$

$\csc A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$

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