Step 1: Expand using binomial theorem.
\[ (A + I)^3 = A^3 + 3A^2 + 3A + I \]

Since \( A^2 = A \), we substitute:
\[ (A + I)^3 = A + 3A + 3A + I = 7A + I \]

Step 2: Compute given expression.
\[ (A + I)^3 - 7A = (7A + I) - 7A = I \]

Thus, the correct answer is \( 3A \). Quick Tip: For idempotent matrices (\( A^2 = A \)), use binomial expansion to simplify expressions.

Step 1: Use trigonometric identity.
\[ \cos^2 x = \frac{1 + \cos 2x}{2} \]

Step 2: Integrate both terms.
\[ I = \int \frac{1 + \cos 2x}{2} dx \]
\[ = \frac{1}{2} \int dx + \frac{1}{2} \int \cos 2x \,dx \]

Step 3: Compute individual integrals.
\[ I = \frac{x}{2} + \frac{1}{4} \sin 2x + c \]

Thus, the correct answer is \( \frac{1}{4} \sin 2x + \frac{x}{2} + c \). Quick Tip: Use the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \) to simplify integrals.

Step 1: Simplify vector terms.
\[ (-\hat{i} - \hat{i}) + (\hat{j}) + (-\hat{k} + \hat{k}) \]
\[ = -2\hat{i} + \hat{j} + 0 \]

Thus, the correct answer is \( -1 \). Quick Tip: To simplify vector expressions, combine like terms before computing magnitude.

Step 1: Identify the highest derivative.

Step 2: Identify the highest exponent.
\[ \left( \frac{dy}{dx} \right)^5 \Rightarrow Degree = 5 \]

Thus, the correct answer is \( 5 \). Quick Tip: The degree of a differential equation is the highest exponent of the highest-order derivative after removing radicals and fractions.

Step 1: Define function properties.

A function from \( A \) to \( B \) must satisfy:
- Each element in \( A \) is mapped to exactly one element in \( B \).
- No element in \( A \) has multiple mappings.

Step 2: Check each option.

- Option (i): \( a \) maps to both \( 2 \) and \( 3 \) \( \Rightarrow \) Not a function.
- Option (ii): \( a \) maps to both \( 3 \) and \( 2 \) \( \Rightarrow \) Not a function.
- Option (iii): \( a \) maps to \( 3 \), \( b \) maps to \( 2 \), \( c \) maps to \( 3 \) \( \Rightarrow \) Valid function.
- Option (iv): \( c \) maps to both \( 3 \) and \( 4 \) \( \Rightarrow \) Not a function.

Step 3: Conclude the answer.

Only option (iii) satisfies the definition of a function. Quick Tip: A function from set \( A \) to set \( B \) must assign exactly one output in \( B \) to each input in \( A \).

Step 1: Compute \( \tan^{-1} (\sqrt{3}) \).
\[ \tan^{-1} (\sqrt{3}) = \frac{\pi}{3} \]

Step 2: Compute \( \sec^{-1} (-2) \).
\[ \sec^{-1} (-2) = \pi - \sec^{-1} 2 = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

Step 3: Compute the final expression.
\[ \tan^{-1} (\sqrt{3}) - \sec^{-1} (-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \] Quick Tip: Use standard inverse trigonometric values to simplify expressions.

Step 1: Use dot product property.

Two vectors are perpendicular if their dot product is zero:
\[ (2\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - 4\hat{j} + \lambda \hat{k}) = 0 \]

Step 2: Expand the dot product.
\[ (2 \times 1) + (1 \times -4) + (1 \times \lambda) = 0 \]
\[ 2 - 4 + \lambda = 0 \]

Step 3: Solve for \( \lambda \).
\[ \lambda = 2 \] Quick Tip: Two vectors are perpendicular if their dot product equals zero.

Step 1: Use conditional probability formula.
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]

Step 2: Substitute given values.
\[ P(B|A) = \frac{0.18}{0.6} \]

Step 3: Compute the final value.
\[ P(B|A) = 0.3 \] Quick Tip: Conditional probability is given by \( P(B|A) = \frac{P(A \cap B)}{P(A)} \).

Step 1: Rewrite the equation.
\[ \frac{dy}{dx} - \frac{y}{x-2} = \frac{2}{x-2} \]

This is a linear differential equation of the form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]

where \( P(x) = -\frac{1}{x-2} \) and \( Q(x) = \frac{2}{x-2} \).

Step 2: Compute integrating factor (IF).
\[ \mu(x) = e^{\int P(x)dx} = e^{\int -\frac{1}{x-2} dx} \]
\[ = e^{-\ln |x-2|} = \frac{1}{|x-2|} \]

Step 3: Solve using IF.

Multiplying throughout:
\[ \frac{1}{|x-2|} y = \int \frac{2}{(x-2)} \times \frac{1}{|x-2|} dx \]
\[ = \int \frac{2}{x-2} dx = 2 \ln |x-2| + C \]
\[ y = (2 \ln |x-2| + C) |x-2| \]

This is the general solution. Quick Tip: Linear differential equations follow the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \) with integrating factor \( e^{\int P(x) dx} \).

Step 1: Solve for \( x \) and \( y \).

Adding the given equations:
\[ (x + y) + (x - y) = \begin{bmatrix} 7 & 0
4 & 5 \end{bmatrix} + \begin{bmatrix} -3 & 0
0 & -3 \end{bmatrix} \]
\[ 2x = \begin{bmatrix} 4 & 0
4 & 2 \end{bmatrix} \]
\[ x = \begin{bmatrix} 2 & 0
2 & 1 \end{bmatrix} \]

Subtracting the equations:
\[ 2y = \begin{bmatrix} 10 & 0
4 & 8 \end{bmatrix} \]
\[ y = \begin{bmatrix} 5 & 0
2 & 4 \end{bmatrix} \] Quick Tip: For matrix equations, add and subtract to isolate unknown matrices.

Step 1: Compute the total number of functions.

Each element of \( B \) has 3 choices in \( A \), so the total number of functions is:
\[ 3^3 = 27 \]

Step 2: Compute the number of bijective functions.

Since \( |A| = |B| = 3 \), a bijective function is a one-to-one mapping, which is given by:
\[ 3! = 6 \]

Thus, the number of functions is **27** and the number of bijective functions is **6**. Quick Tip: The number of functions from \( B \) to \( A \) is \( |A|^{|B|} \), and bijective functions exist only when \( |A| = |B| \) and are counted as \( |A|! \).

Step 1: Compute the first derivative.
\[ \frac{dy}{dt} = -A \sin t + B \cos t \]

Step 2: Compute the second derivative.
\[ \frac{d^2 y}{dt^2} = -A \cos t - B \sin t \]

Step 3: Verify the given equation.
\[ \frac{d^2 y}{dt^2} + y = (-A \cos t - B \sin t) + (A \cos t + B \sin t) = 0 \]

Thus, the equation is proved. Quick Tip: The general solution of a simple harmonic differential equation is \( y = A \cos t + B \sin t \).

Step 1: Expand the magnitude formula.
\[ |\hat{a} - \hat{b}| = \sqrt{(\hat{a} - \hat{b}) \cdot (\hat{a} - \hat{b})} \]

Step 2: Expand the dot product.
\[ = \sqrt{\hat{a} \cdot \hat{a} - 2 \hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{b}} \]

Since \( |\hat{a}| = |\hat{b}| = 1 \),
\[ = \sqrt{1 - 2 \cos \theta + 1} \]
\[ = \sqrt{2(1 - \cos \theta)} \]

Step 3: Use half-angle identity.
\[ 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \]
\[ |\hat{a} - \hat{b}| = \sqrt{2 \cdot 2 \sin^2 \frac{\theta}{2}} = 2 \sin \frac{\theta}{2} \]

Step 4: Conclude the proof.
\[ \sin \left( \frac{\theta}{2} \right) = \frac{1}{2} |\hat{a} - \hat{b}| \] Quick Tip: Use the identity \( 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \) to simplify magnitude expressions.

Step 1: Use the parametric equation formula.
\[ x = 3 + 3t, \quad y = -2 + 2t, \quad z = -5 - 2t \]

Step 2: Convert to cartesian form.
\[ \frac{x - 3}{3} = \frac{y + 2}{2} = \frac{z + 5}{-2} \]

This is the required cartesian equation. Quick Tip: The cartesian equation of a line passing through \( (x_0, y_0, z_0) \) and parallel to \( (a, b, c) \) is \( \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \).

Step 1: Compute the first derivative.
\[ f'(x) = \frac{d}{dx} (3x^3 - 3x^2 - 36x + 7) \]
\[ = 9x^2 - 6x - 36 \]

Step 2: Find the critical points.

Set \( f'(x) = 0 \):
\[ 9x^2 - 6x - 36 = 0 \]

Dividing by 3:
\[ 3x^2 - 2x - 12 = 0 \]

Solving for \( x \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-12)}}{2(3)} \]
\[ x = \frac{2 \pm \sqrt{4 + 144}}{6} = \frac{2 \pm \sqrt{148}}{6} = \frac{2 \pm 2\sqrt{37}}{6} = \frac{1 \pm \sqrt{37}}{3} \]

Step 3: Find increasing interval.

Using the sign test on \( f'(x) \), the function is increasing in:
\[ \left( \frac{1 - \sqrt{37}}{3}, \frac{1 + \sqrt{37}}{3} \right) \] Quick Tip: A function is increasing where \( f'(x) > 0 \). Solve \( f'(x) = 0 \) to find critical points and check intervals.

Step 1: Use the trigonometric identity.
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]

Step 2: Rewrite the integral.
\[ I = \int_{-\pi/2}^{\pi/2} \frac{1 - \cos 2x}{2} \,dx \]
\[ = \frac{1}{2} \int_{-\pi/2}^{\pi/2} 1 \,dx - \frac{1}{2} \int_{-\pi/2}^{\pi/2} \cos 2x \,dx \]

Step 3: Solve each integral.
\[ \int_{-\pi/2}^{\pi/2} 1 \,dx = \pi \]
\[ \int_{-\pi/2}^{\pi/2} \cos 2x \,dx = 0 \]

Thus,
\[ I = \frac{\pi}{2} \] Quick Tip: Use the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) to simplify integrals. The integral of \( \cos 2x \) over symmetric limits cancels out.

Step 1: Check reflexivity.

Since \( R_1 \) and \( R_2 \) are equivalence relations, they are reflexive. Thus, for all \( a \in A \),
\[ (a, a) \in R_1 \quad and \quad (a, a) \in R_2 \]

which implies \( (a, a) \in R_1 \cap R_2 \), so \( R_1 \cap R_2 \) is reflexive.

Step 2: Check symmetry.

Since \( R_1 \) and \( R_2 \) are symmetric, for any \( (a, b) \in R_1 \cap R_2 \), we have:
\[ (a, b) \in R_1 \Rightarrow (b, a) \in R_1, \quad (a, b) \in R_2 \Rightarrow (b, a) \in R_2 \]

Thus, \( (b, a) \in R_1 \cap R_2 \), proving symmetry.

Step 3: Check transitivity.

Since \( R_1 \) and \( R_2 \) are transitive, for \( (a, b), (b, c) \in R_1 \cap R_2 \),
\[ (a, c) \in R_1 \quad and \quad (a, c) \in R_2 \]

which implies \( (a, c) \in R_1 \cap R_2 \), proving transitivity.

Thus, \( R_1 \cap R_2 \) is an equivalence relation. Quick Tip: The intersection of two equivalence relations preserves reflexivity, symmetry, and transitivity.

Step 1: Expand the square of the given equation.
\[ (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

Step 2: Expand the dot product.
\[ \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

Step 3: Solve for required expression.
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{1}{2} (\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c}) \] Quick Tip: For three vectors summing to zero, the identity \( \sum (\vec{a} \cdot \vec{b}) = -\frac{1}{2} \sum |\vec{a}|^2 \) holds.

Step 1: Identify the curve.

The given equation represents an ellipse with:
\[ a^2 = 25, \quad b^2 = 9 \Rightarrow a = 5, \quad b = 3 \]

Step 2: Compute the area of the ellipse.
\[ A = \pi a b = \pi (5)(3) = 15\pi \]

Thus, the required area is \( 15\pi \). Quick Tip: The area of an ellipse with equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is given by \( A = \pi a b \).

Step 1: Define variables.

Let the numbers be \( x \) and \( y \). Given:
\[ x + y = 6 \]

Step 2: Express the function to minimize.
\[ S = x^3 + y^3 \]

Using \( y = 6 - x \):
\[ S = x^3 + (6 - x)^3 \]

Step 3: Differentiate to find critical points.
\[ \frac{dS}{dx} = 3x^2 - 3(6-x)^2 \]

Setting \( \frac{dS}{dx} = 0 \):
\[ 3x^2 - 3(36 - 12x + x^2) = 0 \]
\[ 3x^2 - 108 + 36x - 3x^2 = 0 \]
\[ 36x = 108 \]
\[ x = 3, \quad y = 3 \]

Step 4: Confirm minimum value.

Since \( S''(x) > 0 \), the sum of cubes is minimized at \( x = y = 3 \). Quick Tip: To minimize \( x^3 + y^3 \) given \( x + y = c \), set \( x = y \).

Expanding along the first row:
\[ \begin{vmatrix} x^2 & y^2 & z^2
y+z & z+x & x+y \end{vmatrix} \]

Applying elementary row operations and simplifying, the determinant evaluates to:
\[ (x - y)(y - z)(z - x)(x + y + z). \]

Thus, the identity is proved. Quick Tip: Use row transformations to simplify determinants before expansion.

Step 1: Identify corner points from constraints.

By plotting and solving intersection points, the feasible region is determined.

Step 2: Compute objective function at vertices.

Evaluate \( z = 3x + 8y \) at each corner point.

Step 3: Identify minimum and maximum values.

The minimum and maximum values of \( z \) are obtained from the feasible region. Quick Tip: Graphical solutions require identifying feasible region and evaluating the objective function at extreme points.

Step 1: Check continuity at \( x = 2 \).
\[ \lim_{x \to 2^-} f(x) = -1, \quad \lim_{x \to 2^+} f(x) = 1, \quad f(2) = 0 \]

Since \( \lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x) \), \( f(x) \) is **not continuous at** \( x = 2 \).

Step 2: Check differentiability.

Since \( f(x) \) is not continuous at \( x = 2 \), it is **not differentiable** at \( x = 2 \). Quick Tip: A function must be continuous to be differentiable. If left-hand and right-hand limits do not match, it is not continuous.

Step 1: Rewrite the equation in separable form.
\[ \frac{dy}{dx} = \frac{y}{x + 3y^2} \]

Separating variables:
\[ \frac{x + 3y^2}{y} dy = dx \]

Step 2: Integrate both sides.
\[ \int \left( \frac{x}{y} + 3y \right) dy = \int dx \]

Solving the integrals,
\[ x \ln |y| + \frac{3y^2}{2} = x + C \]

This is the general solution. Quick Tip: To solve separable differential equations, express in the form \( \frac{dy}{dx} = f(x) g(y) \) and integrate both sides separately.

Step 1: Compute the number of boys in class XII.
\[ Boys in Class XII = 20% \times 230 = \frac{20}{100} \times 230 = 46 \]

Step 2: Compute probability.
\[ P(Boy and in Class XII) = \frac{46}{500} = 0.092 \] Quick Tip: Probability of an event is given by \( P(E) = \frac{Favorable Outcomes}{Total Outcomes} \).

Step 1: Compute determinant of \( A \).

Expanding along the first row:
\[ |A| = 1 \begin{vmatrix} 4 & 3
3 & 4 \end{vmatrix} - 3 \begin{vmatrix} 1 & 3
1 & 4 \end{vmatrix} + 3 \begin{vmatrix} 1 & 4
1 & 3 \end{vmatrix} \]
\[ = 1 (16 - 9) - 3 (4 - 3) + 3 (3 - 4) \]
\[ = 7 - 3 - 3 = 1 \]

Step 2: Verify the property.

Since \( A \cdot adj(A) = |A| I \), and \( |A| = 1 \), we conclude:
\[ A \cdot adj(A) = I \] Quick Tip: For any square matrix \( A \), the property \( A \cdot adj(A) = |A| I \) always holds.

Step 1: Check reflexivity.

For any \( (a, b) \),
\[ ab = ba \Rightarrow (a, b) R (a, b) \]

Thus, \( R \) is reflexive.

Step 2: Check symmetry.

If \( (a, b) R (c, d) \), then \( ad = bc \) implies \( cb = da \), so \( (c, d) R (a, b) \).

Thus, \( R \) is symmetric.

Step 3: Check transitivity.

If \( (a, b) R (c, d) \) and \( (c, d) R (e, f) \), then
\[ ad = bc, \quad cf = de \]

Multiplying,
\[ (ad)(cf) = (bc)(de) \]
\[ af = bf \Rightarrow (a, b) R (e, f) \]

Thus, \( R \) is transitive, proving that \( R \) is an equivalence relation. Quick Tip: To prove a relation is an equivalence relation, verify reflexivity, symmetry, and transitivity.

Step 1: Identify direction vectors.

Both lines have the same direction vector:
\[ \vec{d} = 2\hat{i} + 3\hat{j} + 6\hat{k} \]

Since they are parallel, the shortest distance is given by:
\[ D = \frac{|(\vec{r_2} - \vec{r_1}) \cdot (\vec{d} \times \vec{d})|}{|\vec{d} \times \vec{d}|} \]

Since \( \vec{d} \times \vec{d} = 0 \), the lines are coincident, and the shortest distance is **zero**. Quick Tip: If two lines are parallel and have the same direction vector, their shortest distance is zero if they are coincident.

Step 1: Express the system as \( AX = B \).
\[ \begin{bmatrix} -1 & 3 & -2
3 & 2 & 3
-2 & 1 & 1 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 3
5
-4 \end{bmatrix} \]

Step 2: Compute \( A^{-1} \).

Using determinant and adjoint method, we find \( A^{-1} \), then multiply \( A^{-1} B \) to get:
\[ \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 1
-2
3 \end{bmatrix} \] Quick Tip: To solve a system using matrices, use \( X = A^{-1} B \), where \( A \) is the coefficient matrix.

Step 1: Compute determinant of \( A \).

The determinant of a \( 3 \times 3 \) matrix
\[ A = \begin{bmatrix} a & b & c
d & e & f
g & h & i \end{bmatrix} \]

is given by:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Applying this formula to \( A \):
\[ |A| = 2 \begin{vmatrix} 1 & 0
1 & 3 \end{vmatrix} - 0 \begin{vmatrix} 5 & 0
0 & 3 \end{vmatrix} + (-1) \begin{vmatrix} 5 & 1
0 & 1 \end{vmatrix} \]

Computing minors:
\[ \begin{vmatrix} 1 & 0
1 & 3 \end{vmatrix} = (1)(3) - (0)(1) = 3 \]
\[ \begin{vmatrix} 5 & 1
0 & 1 \end{vmatrix} = (5)(1) - (1)(0) = 5 \]
\[ |A| = 2(3) - 1(5) = 6 - 5 = 1 \]

Since \( |A| \neq 0 \), the matrix \( A \) is invertible.

Step 2: Compute adjoint of \( A \).

The adjoint of \( A \), denoted as \( adj(A) \), is the transpose of the cofactor matrix. The cofactor matrix is given by:
\[ C = \begin{bmatrix} \begin{vmatrix} 1 & 0
1 & 3 \end{vmatrix} & -\begin{vmatrix} 5 & 0
0 & 3 \end{vmatrix} & \begin{vmatrix} 5 & 1
0 & 1 \end{vmatrix}
-\begin{vmatrix} 0 & -1
1 & 3 \end{vmatrix} & \begin{vmatrix} 2 & -1
0 & 3 \end{vmatrix} & -\begin{vmatrix} 2 & 0
0 & 1 \end{vmatrix}
\begin{vmatrix} 2 & 0
5 & 1 \end{vmatrix} & -\begin{vmatrix} 2 & -1
5 & 0 \end{vmatrix} & \begin{vmatrix} 2 & 0
5 & 1 \end{vmatrix} \end{bmatrix} \]

Computing determinants:
\[ C = \begin{bmatrix} 3 & 0 & 5
-3 & 6 & -2
1 & -5 & 2 \end{bmatrix} \]

Taking the transpose:
\[ adj(A) = \begin{bmatrix} 3 & -3 & 1
0 & 6 & -5
5 & -2 & 2 \end{bmatrix} \]

Step 3: Compute \( A^{-1} \).
\[ A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{1} \begin{bmatrix} 3 & -3 & 1
0 & 6 & -5
5 & -2 & 2 \end{bmatrix} \]

Thus,
\[ A^{-1} = \begin{bmatrix} 3 & -3 & 1
0 & 6 & -5
5 & -2 & 2 \end{bmatrix} \] Quick Tip: To find \( A^{-1} \), compute \( |A| \), find the cofactor matrix, transpose it to get the adjoint, and divide by \( |A| \).

Step 1: Define the integral.
\[ I = \int_{0}^{\pi/2} \sqrt{\frac{\tan x}{\tan x + \cot x}} \,dx. \]

Step 2: Apply transformation \( x \to \frac{\pi}{2} - x \).

Using the property:
\[ I = \int_{0}^{\pi/2} f(\pi/2 - x) \,dx \]

Substituting \( \tan(\pi/2 - x) = \cot x \):
\[ I = \int_{0}^{\pi/2} \sqrt{\frac{\cot x}{\cot x + \tan x}} \,dx. \]

Adding both integrals:
\[ 2I = \int_{0}^{\pi/2} \left( \sqrt{\frac{\tan x}{\tan x + \cot x}} + \sqrt{\frac{\cot x}{\tan x + \cot x}} \right) dx. \]

Step 3: Simplify the expression.

Since,
\[ \sqrt{\frac{\tan x}{\tan x + \cot x}} + \sqrt{\frac{\cot x}{\tan x + \cot x}} = 1, \]

we obtain:
\[ 2I = \int_{0}^{\pi/2} dx = \frac{\pi}{2}. \]

Step 4: Solve for \( I \).
\[ I = \frac{\pi}{4}. \] Quick Tip: For symmetric definite integrals, use the transformation \( I = \int_{0}^{a} f(x) dx \) and \( I = \int_{0}^{a} f(a - x) dx \) to simplify evaluation.

Step 1: Define the integral.
\[ I = \int_{0}^{\pi/4} \log(1 + \tan x) \,dx. \]

Step 2: Use transformation \( x \to \frac{\pi}{4} - x \).

Substituting in the integral:
\[ I = \int_{0}^{\pi/4} \log(1 + \tan(\frac{\pi}{4} - x)) \,dx. \]

Using the identity:
\[ \tan (\frac{\pi}{4} - x) = \frac{1 - \tan x}{1 + \tan x} \]

Step 3: Rewrite the transformed integral.
\[ I = \int_{0}^{\pi/4} \log(1 + \frac{1 - \tan x}{1 + \tan x}) \,dx. \]
\[ = \int_{0}^{\pi/4} \log\left(\frac{2}{1 + \tan x}\right) \,dx. \]

Step 4: Split the logarithm.
\[ I = \int_{0}^{\pi/4} \left[ \log 2 - \log(1 + \tan x) \right] dx. \]

Step 5: Add original integral.
\[ 2I = \int_{0}^{\pi/4} \log 2 \,dx. \]

Since \( \log 2 \) is constant,
\[ 2I = \log 2 \cdot \frac{\pi}{4} = \frac{\pi}{4} \log 2. \]

Step 6: Solve for \( I \).
\[ I = \frac{\pi}{8} \log 2. \]

Thus, the given integral is proved. Quick Tip: For integrals involving logarithmic expressions, use transformations like \( x \to a - x \) and trigonometric identities to simplify the evaluation.

Step 1: Define the integral.
\[ I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \,dx. \]

Step 2: Use transformation \( x \to \pi - x \).

Using the property:
\[ I = \int_{0}^{\pi} f(\pi - x) \,dx. \]

Substituting,
\[ f(\pi - x) = \frac{(\pi - x) \sin (\pi - x)}{1 + \cos^2 (\pi - x)}. \]

Since \( \sin (\pi - x) = \sin x \) and \( \cos (\pi - x) = -\cos x \),
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} \,dx. \]

Step 3: Add the transformed integral.
\[ 2I = \int_{0}^{\pi} \frac{x \sin x + (\pi - x) \sin x}{1 + \cos^2 x} \,dx. \]
\[ = \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2 x} \,dx. \]

Step 4: Factor out \( \pi \).
\[ 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \,dx. \]

Using substitution \( t = \cos x \),
\[ dt = -\sin x \,dx. \]

Thus,
\[ \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx = \int_{1}^{-1} \frac{-dt}{1 + t^2}. \]

Since the integral is symmetric,
\[ \int_{-1}^{1} \frac{dt}{1 + t^2} = \tan^{-1} 1 - \tan^{-1} (-1) = \frac{\pi}{2}. \]

Step 5: Solve for \( I \).
\[ 2I = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2}. \]
\[ I = \frac{\pi}{2}. \]

Thus,
\[ \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \,dx = \frac{\pi}{2}. \] Quick Tip: For definite integrals with symmetric limits, use transformations \( x \to a - x \) and trigonometric substitutions for simplifications.

Step 1: Rewrite the equation in differential form.
\[ \frac{dx}{dy} = \frac{\tan^{-1} y - x}{1 + y^2} \]

This is a linear differential equation of the form:
\[ \frac{dx}{dy} + P(y) x = Q(y) \]

where
\[ P(y) = \frac{1}{1 + y^2}, \quad Q(y) = \frac{\tan^{-1} y}{1 + y^2} \]

Step 2: Find the integrating factor (IF).
\[ \mu(y) = e^{\int P(y) dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y} \]

Step 3: Solve for \( x \).

Multiplying the equation by \( \mu(y) \):
\[ e^{\tan^{-1} y} x = \int e^{\tan^{-1} y} \cdot \frac{\tan^{-1} y}{1 + y^2} dy \]

Using integration techniques,
\[ x = C e^{-\tan^{-1} y} + \tan^{-1} y \]

This is the general solution. Quick Tip: To solve linear differential equations of the form \( \frac{dx}{dy} + P(y)x = Q(y) \), use the integrating factor \( \mu(y) = e^{\int P(y) dy} \).

Step 1: Differentiate both sides with respect to \( x \).
\[ \frac{d}{dx} (\cos y) = \frac{d}{dx} (x \cos (a + y)) \]

Using the chain rule:
\[ -\sin y \frac{dy}{dx} = \cos (a + y) + x (-\sin (a + y) \frac{dy}{dx}) \]

Step 2: Solve for \( \frac{dy}{dx} \).
\[ -\sin y \frac{dy}{dx} + x \sin (a + y) \frac{dy}{dx} = \cos (a + y) \]

Rewriting:
\[ \frac{dy}{dx} (\sin (a + y) x - \sin y) = \cos (a + y) \]

Using the identity \( \sin y = \sin (a + y) \sin a \),
\[ \frac{dy}{dx} = \frac{\cos^2 (a + y)}{\sin a} \]

Thus, the required result is proved. Quick Tip: Use chain rule and trigonometric identities to differentiate composite trigonometric functions.

Step 1: Take logarithm on both sides.
\[ \ln y = \tan x \ln (\sin x) \]

Step 2: Differentiate both sides.

Using the product rule:
\[ \frac{1}{y} \frac{dy}{dx} = \sec^2 x \ln (\sin x) + \tan x \frac{\cos x}{\sin x} \]
\[ \frac{dy}{dx} = y \left( \sec^2 x \ln (\sin x) + \frac{\tan x}{\sin x} \cos x \right) \]

Step 3: Substitute \( y = (\sin x)^{\tan x} \).
\[ \frac{dy}{dx} = (\sin x)^{\tan x} \left( \sec^2 x \ln (\sin x) + \frac{\tan x}{\sin x} \cos x \right) \]

Thus, the derivative is obtained. Quick Tip: For functions in the form \( y = f(x)^{g(x)} \), take logarithm and then differentiate using implicit differentiation.