Bihar Board Class 10 Mathematics Question Paper 2023 (Code 110 Set-C) Available- Download Here with Solution PDF

Step 1: Use the distance formula from the origin.

For a point \( (x, y) \), distance from origin \( O(0,0) \) is \( \sqrt{x^2 + y^2} \).


Step 2: Substitute \( x = 15 \) and \( y = 8 \).
\[ \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17. \]


Step 3: Conclude.

Hence, the required distance is \( 17 \).
Quick Tip: Distance from the origin is a special case of the distance formula: \( d=\sqrt{(x-0)^2 + (y-0)^2}=\sqrt{x^2+y^2} \).

Step 1: Use the line equation \( y = 2x - 3 \).

A point \( (x, y) \) lies on the line if its coordinates satisfy the equation \( y = 2x - 3 \).


Step 2: Check each option.

For \( (2,2) \): \( 2x-3 = 2\cdot 2 - 3 = 1 \neq 2 \). Not on the line.

For \( (3,4) \): \( 2\cdot 3 - 3 = 3 \neq 4 \). Not on the line.

For \( (4,1) \): \( 2\cdot 4 - 3 = 5 \neq 1 \). Not on the line.

For \( (5,7) \): \( 2\cdot 5 - 3 = 10 - 3 = 7 \). Satisfied.


Step 3: Conclude.

Hence, the line passes through \( (5, 7) \).
Quick Tip: To verify if a point lies on a line, substitute \(x\) and \(y\) into the line's equation and check if the equality holds.

Step 1: Represent the required point on the \(x\)-axis.

Any point on the \(x\)-axis is of the form \( (x, 0) \).


Step 2: Use the equidistant condition (distance formula).

Equidistant from \( (-2,0) \) and \( (6,0) \) means:
\[ \sqrt{(x+2)^2+(0-0)^2} \;=\; \sqrt{(x-6)^2+(0-0)^2}. \]


Step 3: Square and solve for \(x\).
\[ (x+2)^2=(x-6)^2 \;\Rightarrow\; x^2+4x+4=x^2-12x+36 \;\Rightarrow\; 16x=32 \;\Rightarrow\; x=2. \]

Thus the point is \( (2,0) \).
Quick Tip: On a straight line, the point equidistant from two points lies at their midpoint. Here, the midpoint of \(x=-2\) and \(x=6\) is \(x=\frac{-2+6}{2}=2\).

Step 1: Recall the rule.

The perpendicular distance of a point \( (x, y) \) from the \(y\)-axis is \( |x| \).


Step 2: Substitute the coordinates.

For \( (13, 15) \), distance from the \(y\)-axis \(= |13| = 13 \).


Step 3: Conclude.

Hence, the required distance is \( 13 \).
Quick Tip: From the \(y\)-axis, use the absolute value of the \(x\)-coordinate; from the \(x\)-axis, use the absolute value of the \(y\)-coordinate.

Step 1: Use the midpoint formula for the centre.

For endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) of a diameter, the centre is
\[ \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right). \]


Step 2: Substitute the given points.
\[ \left(\frac{-10+6}{2}, \frac{6+(-10)}{2}\right) = \left(\frac{-4}{2}, \frac{-4}{2}\right) = (-2, -2). \]


Step 3: Conclude.

Therefore, the centre of the circle is \( (-2, -2) \).
Quick Tip: For any chord (including the diameter), the midpoint of its endpoints gives the point on the circle’s perpendicular bisector; for a diameter, this midpoint is exactly the centre.

Step 1: Recall centroid formula.

For triangle with vertices \( (x_1,y_1), (x_2,y_2), (x_3,y_3) \), centroid \( G \) is
\[ G\left(\frac{x_1+x_2+x_3}{3},\; \frac{y_1+y_2+y_3}{3}\right). \]


Step 2: Substitute the given coordinates.
\[ G\left(\frac{2+0+4}{3},\; \frac{4+6+(-1)}{3}\right) = \left(\frac{6}{3},\; \frac{9}{3}\right) = (2, 3). \]


Step 3: Conclude.

Thus, the centroid of the triangle is \( (2, 3) \).
Quick Tip: Centroid is the average of the three vertices’ coordinates: add the \(x\)'s and \(y\)'s separately and divide each by \(3\).

Step 1: Recall the rule.

The perpendicular distance of a point \( (x, y) \) from the \(x\)-axis is \( |y| \).


Step 2: Substitute the coordinates.

For \( (13, 19) \), the distance from the \(x\)-axis is \( |19| = 19 \).


Step 3: Conclude.

Hence, the required length of the perpendicular is \( 19 \).
Quick Tip: Distance from the \(x\)-axis depends only on the \(y\)-coordinate: \(d=|y|\). From the \(y\)-axis, use \(d=|x|\).

Step 1: Use the midpoint formula.

Midpoint \(M\) of \( (x_1,y_1) \) and \( (x_2,y_2) \) is \( M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right). \)


Step 2: Substitute the given points.
\[ M\left(\frac{-2+(-6)}{2}, \frac{8+(-4)}{2}\right) = \left(\frac{-8}{2}, \frac{4}{2}\right) = (-4, 2). \]


Step 3: Identify the quadrant.

Since \(x<0\) and \(y>0\), the point \((-4,2)\) lies in the Second quadrant.
Quick Tip: Quadrant signs: I \((+,+)\), II \((-,+)\), III \((-,-)\), IV \((+,-)\). For midpoints, average the coordinates component-wise.

Step 1: Observe the triangle is right-angled on the axes.

Points \(Q(8,0)\) and \(R(0,12)\) lie on the coordinate axes, with \(P(0,0)\) at the origin. Hence, \(PQ\) is along the \(x\)-axis with length \(8\) and \(PR\) is along the \(y\)-axis with length \(12\).


Step 2: Use area formula for a right triangle.
\[ Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times 8 \times 12 = 48. \]


Step 3: Conclude.

Therefore, the area of \( \triangle PQR \) is \(48\).
Quick Tip: When two vertices lie on the axes and the third is the origin, the triangle is right-angled at the origin; use \(\tfrac{1}{2}\times (x-intercept)\times(y-intercept)\).

Step 1: Find the side lengths using the distance formula.
\[ AB: (0,6) to (0,0) \Rightarrow \sqrt{(0-0)^2+(6-0)^2}=6.\] \[BC: (0,0) to (8,0) \Rightarrow \sqrt{(8-0)^2+(0-0)^2}=8.\] \[CA: (8,0) to (0,6) \Rightarrow \sqrt{(8-0)^2+(0-6)^2}=\sqrt{64+36}=10. \]


Step 2: Add the side lengths to get the perimeter.
\[ Perimeter = 6+8+10 = 24. \]


Step 3: Conclude.

Hence, the perimeter of the triangle is \(24\).
Quick Tip: When vertices lie on axes, two sides are axis-aligned (lengths are easy), and the third can often form a Pythagorean triple.

Step 1: Understand ``coincident lines.''

Two lines are coincident if they lie exactly on top of each other; every point on one line is also on the other.


Step 2: Interpret in terms of solutions.

A solution to a pair of linear equations corresponds to a point common to both lines.

Since coincident lines share all points, there are infinitely many common points (solutions).


Step 3: Conclusion.

Therefore, a system represented by coincident lines is consistent and dependent with infinitely many solutions.
Quick Tip: Parallel distinct lines \(\Rightarrow\) no solution; intersecting lines \(\Rightarrow\) one solution; coincident lines \(\Rightarrow\) infinitely many solutions.

Step 1: Convert each equation to slope–intercept form \(y=mx+c\).

For \(2x - y - 3 = 0 \Rightarrow y = 2x - 3\), slope \(m_1 = 2\).

For \(12x + 7y - 5 = 0 \Rightarrow 7y = -12x + 5 \Rightarrow y = -\dfrac{12}{7}x + \dfrac{5}{7}\), slope \(m_2 = -\dfrac{12}{7}\).


Step 2: Compare slopes.

Since \(m_1 \neq m_2\) (\(2 \neq -\dfrac{12}{7}\)), the lines are neither parallel nor coincident; hence they intersect.


Step 3: Conclude.

Therefore, the two given lines are intersecting straight lines.
Quick Tip: In 2D, two non-vertical lines: equal slopes \(\Rightarrow\) parallel; equal slopes and equal intercepts \(\Rightarrow\) coincident; unequal slopes \(\Rightarrow\) intersecting.

Step 1: Rearrange the equation.
\(x - y = 0 \;\Rightarrow\; y = x.\) This is a line with slope \(1\).


Step 2: Identify intercept.

When \(x=0\), \(y=0\). Hence the line passes through \((0,0)\), i.e., the origin.


Step 3: Conclude.

Therefore, the graph is a straight line through the origin (making a \(45^\circ\) angle with the positive \(x\)-axis).
Quick Tip: If a line can be written as \(y=mx\) (no constant term), it always passes through the origin. A constant \(c\) in \(y=mx+c\) shifts it off the origin.

Step 1: Test each option by bringing all terms to one side and checking the highest power of \(x\).


(1) \((x+2)(x-2)=x^2-4x^3 \Rightarrow x^2-4=x^2-4x^3 \Rightarrow 4x^3-4=0\).

Highest degree \(=3\) (cubic) \(\Rightarrow\) Not quadratic.


(2) \((x+2)^2=3(x+4) \Rightarrow x^2+4x+4=3x+12 \Rightarrow x^2+x-8=0\).

Highest degree \(=2\) (quadratic) \(\Rightarrow\) Quadratic.


(3) \((2x^2+3)=(5+x)(2x^2-3)\). RHS contains \(x\cdot 2x^2=2x^3\) \(\Rightarrow\) degree \(3\). After simplifying, equation is cubic \(\Rightarrow\) Not quadratic.


(4) \(2x+\dfrac{1}{2x}=4x^2\). Multiply by \(2x\): \(4x^2+1=8x^3 \Rightarrow 8x^3-4x^2-1=0\).

Highest degree \(=3\) (cubic) \(\Rightarrow\) Not quadratic.


Step 2: Conclude.

Only option \((2)\) reduces to a degree-\(2\) equation, hence it is quadratic.
Quick Tip: To identify a quadratic, first clear denominators if any, expand, and collect like terms. If the highest power of \(x\) is exactly \(2\), it’s quadratic.

Step 1: Use the fact that a root satisfies the equation.

If \(x=-3\) is a root, substitute \(x=-3\) into \(2x^2 + px - 3 = 0\):
\[ 2(-3)^2 + p(-3) - 3 = 0 \;\Rightarrow\; 18 - 3p - 3 = 0. \]


Step 2: Solve for \(p\).
\[ 15 - 3p = 0 \;\Rightarrow\; 3p = 15 \;\Rightarrow\; p = 5. \]


Step 3: Conclude.

Therefore, \(p=5\).
Quick Tip: When a value \(r\) is a root of \(ax^2+bx+c=0\), substituting \(x=r\) gives an equation in the unknown parameter(s). Alternatively, you can use the factor theorem \(a(r)^2+b(r)+c=0\).

Step 1: Use the discriminant condition for equal real roots.

For \(ax^2+bx+c=0\), roots are real and equal when \(D=b^2-4ac=0\).

Here \(a=9\), \(b=3k\), \(c=4\).


Step 2: Set the discriminant to zero and solve for \(k\).
\[ D=(3k)^2-4\cdot 9\cdot 4 = 9k^2-144=0 \;\Rightarrow\; 9k^2=144 \;\Rightarrow\; k^2=16 \;\Rightarrow\; k=\pm 4. \]


Step 3: Conclude.

Hence, the required values of \(k\) are \(\boxed{\pm 4}\).
Quick Tip: “Real and equal roots” \(\Rightarrow\) discriminant \(D=0\); “real and distinct” \(\Rightarrow D>0\); “no real roots” \(\Rightarrow D<0\).

Step 1: Use Vieta's formulas for the quadratic \(x^2 + 3px + 2p^2 = 0\).

Sum of roots: \( \alpha + \beta = -3p \).

Product of roots: \( \alpha\beta = 2p^2 \).


Step 2: Express \( \alpha^2 + \beta^2 \) in terms of \(p\).
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-3p)^2 - 2(2p^2) = 9p^2 - 4p^2 = 5p^2. \]


Step 3: Use the given condition \( \alpha^2 + \beta^2 = 5 \).
\[ 5p^2 = 5 \;\Rightarrow\; p^2 = 1 \;\Rightarrow\; p = \pm 1. \]


Step 4: Conclude.

Hence, the required values of \(p\) are \( \boxed{\pm 1} \).
Quick Tip: For a monic quadratic \(x^2+bx+c=0\), use Vieta: \(\alpha+\beta=-b\), \(\alpha\beta=c\). Also, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\).

Step 1: Recognize the equation as a difference of squares.
\(a^{2}p^{2}x^{2}-q^{2} = (apx)^{2}-q^{2} = (apx-q)(apx+q).\)


Step 2: Set each factor equal to zero and solve for \(x\).

From \(apx-q=0 \Rightarrow x=\dfrac{q}{ap}\).

From \(apx+q=0 \Rightarrow x=-\dfrac{q}{ap}\).


Step 3: Conclude.

Hence, the roots are \(x=\pm\dfrac{q}{ap}\).
Quick Tip: Whenever you see \(A^{2}-B^{2}=0\), factor it as \((A-B)(A+B)=0\) to get roots quickly.

Step 1: Use Vieta's formulas for \(ax^2+bx+c=0\).

Sum of roots \(= -\dfrac{b}{a}\), product of roots \(=\dfrac{c}{a}\). For \(x^2-15x+50=0\), we have \(a=1, b=-15, c=50\).


Step 2: Compute sum and product.

Sum \(= -\dfrac{-15}{1}=15\), Product \(=\dfrac{50}{1}=50\).


Step 3: Form and simplify the ratio.
\(Sum : Product = 15 : 50 = \dfrac{15}{5} : \dfrac{50}{5} = 3 : 10.\)
Quick Tip: For a monic quadratic \(x^2+bx+c=0\), sum of roots \(= -b\) and product \(= c\) — no solving required.

Step 1: Use sum and product of roots for a monic quadratic.

If roots are \( \alpha = -5 \) and \( \beta = -1 \), then
\[ \alpha+\beta = -6, \quad \alpha\beta = 5. \]


Step 2: Form the quadratic using \(x^2 - (\alpha+\beta)x + \alpha\beta = 0\).
\[ x^2 - (-6)x + 5 = 0 \;\Rightarrow\; x^2 + 6x + 5 = 0. \]


Step 3: Conclude.

Hence, the required equation is \( x^2 + 6x + 5 = 0 \).
Quick Tip: Given roots \(\alpha,\beta\), the monic quadratic is \(x^2-(\alpha+\beta)x+\alpha\beta=0\). This avoids expanding factors each time.

Step 1: Use the exact value of \(\cos 60^\circ\).
\(\cos 60^\circ = \dfrac{1}{2}\).


Step 2: Substitute and simplify.
\[ \dfrac{\cos 60^\circ + 1}{\cos 60^\circ - 1} = \dfrac{\tfrac{1}{2}+1}{\tfrac{1}{2}-1} = \dfrac{\tfrac{3}{2}}{-\tfrac{1}{2}} = -3. \]


Step 3: Conclude.

Hence, the required value is \(-3\).
Quick Tip: Memorize \(\cos 60^\circ=\tfrac{1}{2}\) and \(\sin 30^\circ=\tfrac{1}{2}\); they simplify many trig ratios instantly.

Step 1: Find angle \(A\).
\(\dfrac{A}{5}=12^\circ \Rightarrow A=60^\circ.\)


Step 2: Evaluate \( \csc^2 A \).
\(\sin 60^\circ=\dfrac{\sqrt{3}}{2} \Rightarrow \csc 60^\circ=\dfrac{1}{\sin 60^\circ}=\dfrac{2}{\sqrt{3}}.\)

Hence, \(\csc^2 60^\circ=\left(\dfrac{2}{\sqrt{3}}\right)^2=\dfrac{4}{3}.\)


Step 3: Multiply by 3.
\(3\csc^2 60^\circ=3\cdot\dfrac{4}{3}=4.\)
Quick Tip: Convert the given angle first; with special angles like \(30^\circ, 45^\circ, 60^\circ\), use exact trig values to simplify quickly.

Step 1: Use the co-function identity.
\(\sin(90^\circ-\theta)=\cos\theta\). With \(\theta=36^\circ\), we get \(\sin 54^\circ=\cos 36^\circ\).


Step 2: Substitute and simplify.
\[ \sin 54^\circ - \cos 36^\circ = \cos 36^\circ - \cos 36^\circ = 0. \]
Quick Tip: Remember co-function pairs: \(\sin(90^\circ-\theta)=\cos\theta\) and \(\cos(90^\circ-\theta)=\sin\theta\).

Step 1: Note the factor \(\cos 90^\circ\).

We know \(\cos 90^\circ = 0\).


Step 2: Use zero-product property.

Since the product includes the factor \( \cos 90^\circ = 0 \), the entire product equals \(0\).
Quick Tip: When a product includes any factor equal to \(0\), the whole product is \(0\) — no further computation needed.

Step 1: Use co-function identities.

Since \(53^\circ = 90^\circ - 37^\circ\), we have \(\sin 53^\circ = \cos 37^\circ\).

Thus, \(\dfrac{\cos 37^\circ}{\sin 53^\circ} = \dfrac{\cos 37^\circ}{\cos 37^\circ} = 1.\)


Step 2: Simplify the second term.

Because \(56^\circ = 90^\circ - 34^\circ\), \(\tan 56^\circ = \cot 34^\circ\).

Hence, \(\dfrac{\cot 34^\circ}{\tan 56^\circ} = \dfrac{\cot 34^\circ}{\cot 34^\circ} = 1.\)


Step 3: Add the results.

Total \(= 1 + 1 = 2.\)
Quick Tip: Co-function pairs: \(\sin(90^\circ-\theta)=\cos\theta\), \(\tan(90^\circ-\theta)=\cot\theta\), \(\sec(90^\circ-\theta)=\csc\theta\). They make many evaluations trivial.

Step 1: Recall exact values of \(\sin 45^\circ\) and \(\cos 45^\circ\).

We know \(\sin 45^\circ = \dfrac{1}{\sqrt{2}}\), \(\cos 45^\circ = \dfrac{1}{\sqrt{2}}\).


Step 2: Substitute into the expression.
\[ 2(\sin 45^\circ - \cos 45^\circ) = 2\left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\right) = 2(0) = 0. \]


Step 3: Conclude.

Therefore, the value is \(0\).
Quick Tip: For \(45^\circ\), sine and cosine are equal, so any difference like \(\sin 45^\circ - \cos 45^\circ\) vanishes to zero.

Step 1: Recall the Pythagorean identity in trigonometry.

For any angle \(\theta\) (where defined), \(\;\csc^{2}\theta = 1 + \cot^{2}\theta.\)


Step 2: Substitute into the expression.
\[ \csc^{2}\theta - \cot^{2}\theta = (1 + \cot^{2}\theta) - \cot^{2}\theta = 1. \]


Step 3: Conclude.

Thus, the required value is \(1\).
Quick Tip: Memorize \(\csc^{2}\theta = 1 + \cot^{2}\theta\) and \(\sec^{2}\theta = 1 + \tan^{2}\theta\). They help reduce many expressions instantly.

Step 1: Express \(\cos 48^\circ\) in terms of \(p\).

Given \( \sin 48^\circ = p \) and \(48^\circ\) is acute, so \[ \cos 48^\circ = \sqrt{1-\sin^2 48^\circ} = \sqrt{1-p^2}. \]


Step 2: Use \( \tan \theta = \dfrac{\sin \theta}{\cos \theta} \).
\[ \tan 48^\circ = \frac{\sin 48^\circ}{\cos 48^\circ} = \frac{p}{\sqrt{1-p^2}}. \]


Step 3: Conclude.

Hence, \( \tan 48^\circ = \dfrac{p}{\sqrt{1-p^{2}}}. \)
Quick Tip: For an acute angle, \( \cos \theta = \sqrt{1-\sin^2\theta} \) and \( \tan \theta = \dfrac{\sin\theta}{\cos\theta} \). Always choose the positive square root.

Step 1: Use co-function identity for tangent.

Since \(63^\circ = 90^\circ - 27^\circ\), we have \(\tan 63^\circ = \cot 27^\circ\).


Step 2: Evaluate the product.
\[ \tan 27^\circ \cdot \tan 63^\circ = \tan 27^\circ \cdot \cot 27^\circ = 1. \]


Step 3: Equate to \(\sin A\) and solve.

Given \( \tan 27^\circ \cdot \tan 63^\circ = \sin A \Rightarrow \sin A = 1 \).

Therefore \( A = 90^\circ \) (in the principal range).
Quick Tip: \(\tan(90^\circ-\theta)=\cot\theta\). Hence \(\tan\theta\cdot\tan(90^\circ-\theta)=1\), a handy shortcut for many angle products.

Step 1: Use co-function identities.
\(\sin(90^\circ-\theta)=\cos\theta\) and \(\csc(90^\circ-\theta)=\sec\theta\).


Step 2: Simplify numerator and denominator.

Numerator: \(\sin^{2}(90^\circ-\theta)+\sin^{2}\theta=\cos^{2}\theta+\sin^{2}\theta=1.\)

Denominator: \(\csc^{2}(90^\circ-\theta)-\tan^{2}\theta=\sec^{2}\theta-\tan^{2}\theta=1\) (Pythagorean identity).


Step 3: Conclude.
\[ \frac{1}{1}=1. \]
Quick Tip: Remember \(\sin(90^\circ-\theta)=\cos\theta\) and the identity \(\sec^{2}\theta-\tan^{2}\theta=1\); such pairs often collapse expressions to a constant.

Step 1: Use the circumference formula.

Distance in one revolution \(= 2\pi r\). For \(2\) revolutions, distance \(= 4\pi r\).


Step 2: Substitute \( r=\dfrac{35}{44} \) m and \( \pi=\dfrac{22}{7} \).
\[ Distance = 4\pi r = 4 \times \frac{22}{7} \times \frac{35}{44} = \frac{140}{44}\times \frac{22}{7} = \frac{35}{11}\times \frac{22}{7} = \frac{770}{77} = 10\ m. \]


Step 3: Conclude.

Hence, the distance covered in \(2\) revolutions is \(10\) metres.
Quick Tip: Distance in \(n\) revolutions \(= n \times 2\pi r\). For clean numbers, use \(\pi=\frac{22}{7}\) when \(r\) has factors \(7\) or \(11\).

Step 1: Use the arc length formula for central angle in degrees.

Length of arc \(s\) corresponding to central angle \(\theta^\circ\) is \[ s=\frac{\theta}{360^\circ}\times 2\pi r. \]


Step 2: Simplify the expression.
\[ s=\frac{\theta}{360}\cdot 2\pi r=\frac{\pi r \theta}{180}. \]


Step 3: Conclude.

Hence, the length of arc \(AB\) is \(\dfrac{\pi r \theta}{180}\).
Quick Tip: For angles in degrees, arc length \(s=\dfrac{\theta}{360}\times 2\pi r\). For angles in radians, \(s=r\theta\).

Step 1: Use volume conservation.

Number of small spheres \(= \dfrac{Volume of big sphere}{Volume of one small sphere}\).


Step 2: Write volumes using \(V=\dfrac{4}{3}\pi r^3\).
\[ N=\frac{\frac{4}{3}\pi (8)^3}{\frac{4}{3}\pi (1)^3} = \frac{512}{1} = 512. \]


Step 3: Conclude.

Thus, \(512\) small spheres can be made.
Quick Tip: When reshaping solids without loss, the count scales with the ratio of volumes. For similar shapes, that’s the cube of the linear scale: \(N=(R/r)^3\).

Step 1: Write the volume formula of a cone.
\(V=\dfrac{1}{3}\pi r^2 h.\)


Step 2: Convert diameter to radius.

Diameter \(=10\) cm \(\Rightarrow\) radius \(r=\dfrac{10}{2}=5\) cm; height \(h=12\) cm.


Step 3: Substitute and compute.
\[ V=\frac{1}{3}\pi(5)^2(12)=\frac{1}{3}\pi\cdot25\cdot12 =\frac{300}{3}\pi=100\pi\ cm^3. \]


Step 4: Conclude.

Therefore, the volume is \(100\pi\ cm^3\).
Quick Tip: Always halve the diameter to get the radius before using \(V=\frac{1}{3}\pi r^2 h\).

Step 1: Use the formula for volume of a hollow cylinder.

Volume \(V = \pi h (R^2 - r^2)\), where \(R\) is external radius, \(r\) is internal radius, and \(h\) is length.


Step 2: Substitute the given values.
\(R=4\) cm, \(r=3\) cm, \(h=10\) cm:
\[ V=\pi \times 10 \times (4^2-3^2) = 10\pi \times (16-9) = 10\pi \times 7 = 70\pi\ cm^3. \]


Step 3: Evaluate.

Using \(\pi=\dfrac{22}{7}\): \(V=70\times \dfrac{22}{7}=220\ cm^3.\)
Quick Tip: For a pipe (hollow cylinder), subtract areas first: \( \pi(R^2-r^2) \), then multiply by length \(h\).

Step 1: Use “equal base areas.”

If base areas are equal, their radii are equal. Let the common radius be \(r\).


Step 2: Use “equal curved surface areas.”

Cylinder CSA \(=2\pi r h\).

Cone CSA \(=\pi r \ell\) (where \(\ell\) is the slant height).

Given they are equal: \(2\pi r h=\pi r \ell \Rightarrow \ell=2h.\)


Step 3: Substitute the cylinder’s height.
\(h=2\) m \(\Rightarrow \ell=2\times 2=4\) m.
Quick Tip: Equal base areas \(\Rightarrow\) equal radii. If CSA\(_cyl\) \(=2\pi rh\) equals CSA\(_cone\) \(=\pi r\ell\), then \(\ell=2h\) immediately.

Step 1: Find the side length from volume.

For a cube with side \(a\), volume \(V=a^3\). Given \(a^3=125 \Rightarrow a=\sqrt[3]{125}=5\ cm.\)


Step 2: Write the space diagonal of a cube.

Space diagonal \(d=a\sqrt{3} = 5\sqrt{3}\ cm.\)


Step 3: Form the required ratio (side : diagonal).
\[ a:d = 5 : 5\sqrt{3} = 1 : \sqrt{3}. \]
Quick Tip: Cube facts: \(V=a^3\), face diagonal \(=a\sqrt{2}\), space diagonal \(=a\sqrt{3}\). Ratios often simplify by canceling \(a\).

Step 1: Use the total surface area (TSA) of a hemisphere.

TSA \(= 3\pi r^2\). Given \(3\pi r^2 = 462\).


Step 2: Solve for \(r\).

Taking \(\pi=\dfrac{22}{7}\):
\[ 3\cdot \frac{22}{7}\, r^2 = 462 \;\Rightarrow\; r^2 = \frac{462 \cdot 7}{66} = 49 \;\Rightarrow\; r = 7\ cm. \]


Step 3: Find the diameter.

Diameter \(= 2r = 2 \times 7 = 14\ cm.\)
Quick Tip: For a hemisphere, \(TSA = 2\pi r^2 + \pi r^2 = 3\pi r^2\). Once \(r\) is known, the diameter is simply \(2r\).

Step 1: Recall the volume formula of a cylinder.
\(V=\pi r^2 h\). For two cylinders, \(V_1:V_2=(r_1^2 h_1):(r_2^2 h_2)\).


Step 2: Substitute the given ratios.
\(r_1:r_2=2:3 \Rightarrow r_1^2:r_2^2=4:9\).
\(h_1:h_2=5:3\).


Step 3: Multiply component-wise.
\[ V_1:V_2=(4\times 5):(9\times 3)=20:27. \]
Quick Tip: For similar solids where \(V\propto r^2h\) (cylinders), square the radius ratio and multiply by the height ratio to get the volume ratio.

Step 1: Recall the volume formula of a sphere.
\(V=\dfrac{4}{3}\pi r^{3}\). Thus, volume is proportional to \(r^{3}\).


Step 2: Apply the scaling.

If \(r\) becomes \(3r\), then the new volume \(V'\propto (3r)^{3}=27r^{3}\).


Step 3: Conclude.

Hence, the volume becomes \(27\) times the original.
Quick Tip: For similar 3D shapes, volumes scale with the cube of the linear scale factor: if \(r\to kr\), then \(V\to k^{3}V\).

Step 1: Simplify the given expression.
\(\sqrt{81} = 9\). Hence, \[ \dfrac{\sqrt{81}}{2} = \dfrac{9}{2}. \]


Step 2: Check the nature of the number.
\(\dfrac{9}{2}=4.5\). It is not an integer, but it can be expressed as a ratio of two integers, i.e., \(\dfrac{9}{2}\).


Step 3: Conclusion.

Hence, \(\dfrac{\sqrt{81}}{2}\) is a rational number.
Quick Tip: If a number can be expressed in the form \(\dfrac{p}{q}\) (where \(p, q\) are integers and \(q\neq 0\)), it is a rational number even if it’s not an integer.

Step 1: Factorize \(192\).
\(192 = 64 \times 3 = 2^6 \times 3.\)


Step 2: Read off the exponent of \(2\).

In \(192 = 2^6 \times 3\), the exponent of \(2\) is \(6\).


Step 3: Conclude.

Therefore, the required exponent is \(6\).
Quick Tip: Break numbers into convenient powers: \(192=64\times3=2^6\times3\). Spotting perfect powers speeds up prime factorization.

Step 1: Recall the division algorithm.

It states \(a=bq+r\) with \(0 \le r < b\). Here \(r=0\), so \(a=bq\).


Step 2: Substitute the given values and solve for \(b\).
\(72 = b \cdot 8 \Rightarrow b = \dfrac{72}{8} = 9.\)


Step 3: Conclude.

Therefore, \(b=9\).
Quick Tip: When the remainder \(r=0\), the dividend is exactly divisible: \(a=bq\), so \(b=\dfrac{a}{q}\).

Step 1: Use the terminating-decimal criterion.

A rational number \(\dfrac{p}{q}\) in lowest terms has a terminating decimal expansion iff the prime factorization of \(q\) contains only \(2\)'s and \(5\)'s, i.e., \(q=2^m5^n\).


Step 2: Check each denominator (in lowest terms).
\(\dfrac{2}{15}\): \(15=3\cdot 5\) (contains \(3\)) \(\Rightarrow\) non-terminating.
\(\dfrac{11}{160}\): \(160=2^5\cdot 5\) (only \(2\) and \(5\)) \(\Rightarrow\) terminating.
\(\dfrac{17}{60}\): \(60=2^2\cdot 3\cdot 5\) (contains \(3\)) \(\Rightarrow\) non-terminating.
\(\dfrac{6}{35}\): \(35=5\cdot 7\) (contains \(7\)) \(\Rightarrow\) non-terminating.


Step 3: Conclude.

Only \(\dfrac{11}{160}\) meets the criterion, so it has a terminating decimal expansion.
Quick Tip: Reduce the fraction first. If the denominator has any prime other than \(2\) or \(5\), the decimal is non-terminating (recurring).

Step 1: Use parity of multiples of 8.

Since \(q\in\mathbb{Z}^+\), \(8q\) is divisible by \(8\) and hence is even.


Step 2: Check each expression’s parity.
\(8q+1 = even+1=odd\).
\(8q+4 = even+4=\textbf{even}\).
\(8q+3 = even+3=odd\).
\(8q+7 = even+7=odd\).


Step 3: Conclude.

Only \(8q+4\) is not odd; it is even.
Quick Tip: Even \(+\) even \(=\) even, even \(+\) odd \(=\) odd. Any multiple of \(2\) (like \(8q\)) is even.

Step 1: Represent consecutive even numbers.

Let the two consecutive even numbers be \(2n\) and \(2n+2 = 2(n+1)\).


Step 2: Take out the common factor.
\(\gcd(2n,\,2n+2)=\gcd\big(2n,\,2(n+1)\big)=2\cdot \gcd\big(n,\,n+1\big).\)


Step 3: Use property of consecutive integers.
\(\gcd(n,\,n+1)=1\) (consecutive integers are co-prime). Hence, the HCF is \(2\times 1=2\).
Quick Tip: Consecutive integers are co-prime. For consecutive even numbers \(2n\) and \(2n+2\), a factor \(2\) is always common, so HCF \(=2\).

Step 1: Let \(r\) be rational and \(i\) be irrational.

Assume for contradiction that \(r+i\) is rational.


Step 2: Use closure of rationals under subtraction.

If \(r+i\) were rational, then \(i=(r+i)-r\) would be the difference of two rationals, hence rational—contradiction.


Step 3: Conclude.

Therefore, \(r+i\) must be irrational.
Quick Tip: Rationals are closed under addition/subtraction; irrationals are not. Adding any nonzero rational to an irrational keeps it irrational.

Step 1: Use the relation between HCF, LCM and product.

For two positive integers \(a\) and \(b\):
\[ a \times b = HCF(a,b) \times LCM(a,b). \]


Step 2: Substitute the given values and solve for LCM.
\[ 8670 = 17 \times LCM \;\Rightarrow\; LCM = \frac{8670}{17} = 510. \]


Step 3: Conclude.

Hence, the LCM is \(510\).
Quick Tip: Always remember: \( HCF \times LCM = \) product of the two numbers. It saves time when either HCF or LCM is known.

Step 1: Evaluate each expression under the square root.
\(\sqrt{64+36}=\sqrt{100}=10\) (rational).
\(\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\) (irrational).
\(\sqrt{49+49}=\sqrt{98}=7\sqrt{2}\) (irrational).
\(\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\) (irrational).


Step 2: Conclude.

Only option (1) simplifies to an integer, hence it is rational.
Quick Tip: A square root is rational only if the radicand is a perfect square (e.g., \(100, 144\)). Otherwise, it’s irrational.

Step 1: Apply Euclid’s division algorithm.

From \(130 = 15 \cdot 8 + 10\), the remainder is \(10\).

Next, divide \(15\) by \(10\): \(15 = 10 \cdot 1 + 5\).

Finally, \(10 = 5 \cdot 2 + 0\). The last non-zero remainder is \(5\).


Step 2: Conclude.

Hence, \(\gcd(130,15)=5\).
Quick Tip: In Euclid’s algorithm, the HCF is the last non-zero remainder in the sequence of divisions.

\[ 1-\sin^{4}\theta=(1-\sin^{2}\theta)(1+\sin^{2}\theta)=\cos^{2}\theta(1+\sin^{2}\theta). \] Quick Tip: Use difference of squares and \(1-\sin^2\theta=\cos^2\theta\).

By Basic Proportionality Theorem, \(\dfrac{AX}{XB}=\dfrac{AY}{YC}\). Hence \(AY:YC=2:3\). Quick Tip: When a segment is parallel to a triangle’s side, it divides the other two sides proportionally.

In similar figures, any corresponding lengths are in the ratio of the square root of the area ratio: \(\sqrt{121/64}=11/8\). Quick Tip: Area ratio \(=\) (linear scale factor)\(^2\). Take square roots to switch to lengths.

Side ratio \(=\) perimeter ratio \(=30:20=3:2\). Thus required side \(=18\times\frac{2}{3}=12\) cm. Quick Tip: For similar polygons, perimeters scale exactly as any side.

\(\dfrac{\sqrt{3}}{4}a^{2}=36\sqrt{3}\Rightarrow a^{2}=144\Rightarrow a=12\) cm. Quick Tip: Memorize \(A_{equi}=\frac{\sqrt3}{4}a^2\).

Intersecting circles have exactly two common tangents (both direct). Quick Tip: Disjoint: 4, externally tangent: 3, intersecting: 2, internally tangent: 1, one inside without touch: 0.

\(OT\perp PT\). In right \(\triangle OPT\): \[ OP^{2}=OT^{2}+PT^{2}=7^{2}+24^{2}=49+576=625\Rightarrow OP=25\ cm. \] Quick Tip: Radius to the point of tangency is perpendicular to the tangent—use Pythagoras.

Tangents from the same external point are equal: \(PT_1=PT_2=17\) cm. Quick Tip: Equal tangents from an external point is a standard theorem—instant result.

Total area \(=\pi r^{2}=\frac{22}{7}\cdot 7^{2}=154\ cm^2\).

Major area \(=154-14=140\ cm^2\). Quick Tip: Major sector area \(=\) total circle area \(-\) minor sector area.

\(\pi r^2=154\Rightarrow r^2=154\cdot\frac{7}{22}=49\Rightarrow r=7\) cm.

Diameter \(=2r=14\) cm. Quick Tip: When areas involve \(154\), try \(\pi=\frac{22}{7}\) to spot perfect squares quickly.

Step 1: Recall the discriminant formula.

For \(ax^{2}+bx+c=0\), the discriminant is \(D=b^{2}-4ac\).


Step 2: Identify coefficients.

Here \(a=1,\ b=4,\ c=b\).


Step 3: Substitute values.
\[ D = 4^{2} - 4(1)(b) = 16 - 4b \]
Quick Tip: For quadratic equations in the form \(x^{2}+px+q=0\), the discriminant is \(p^{2}-4q\).

Step 1: Check sequence (1).
\(0.44-0.4=0.04,\;0.444-0.44=0.004\), not constant. Not an A.P.


Step 2: Check sequence (2).
\(11-1=10,\;111-11=100\), not constant. Not an A.P.


Step 3: Check sequence (3).

Differences \(4-2=2,\;8-4=4,\;16-8=8\), not constant. This is a G.P.


Step 4: Check sequence (4).

Differences \(-4-0=-4,\;-8-(-4)=-4,\;-12-(-8)=-4\), constant. Hence this is an A.P.
Quick Tip: A sequence is an A.P. if the difference between consecutive terms is always the same.

Step 1: Simplify radicals.
\(\sqrt{18}=3\sqrt{2},\;\sqrt{50}=5\sqrt{2},\;\sqrt{98}=7\sqrt{2},\;\sqrt{162}=9\sqrt{2}\).


Step 2: Check differences.
\(5\sqrt{2}-3\sqrt{2}=2\sqrt{2},\;7\sqrt{2}-5\sqrt{2}=2\sqrt{2}\). Constant difference.


Thus, \(d=2\sqrt{2}\).
Quick Tip: Always try to simplify square roots to find patterns in A.P. sequences.

Step 1: Identify terms. First term \(a=2\), common difference \(d=3\).


Step 2: Use formula.
\[ a_{n} = a+(n-1)d \]


Step 3: Substitute.
\[ a_{11}=2+(11-1)\cdot 3=2+30=32 \]
Quick Tip: The \(n\)th term of an A.P. is given by \(a+(n-1)d\).

Step 1: Substitute \(n=7\).
\[ a_{7}=8(7)-2=56-2=54 \]
Quick Tip: When the general term is given, just substitute \(n\) to find the required term.

Step 1: Write formulas.
\(a_{30}=a+29d,\;\;a_{20}=a+19d\).


Step 2: Subtract.
\[ a_{30}-a_{20}=(a+29d)-(a+19d)=10d \]


Step 3: Use condition.
\[ 10d=50 \quad \Rightarrow \quad d=5 \]
Quick Tip: In A.P., \(a_{m}-a_{n}=(m-n)d\). This is a shortcut for differences.

Step 1: Use A.P. property.

In an A.P., the middle term equals the average of its neighbors:
\[ 3x+2=\frac{(2x-1)+(6x-1)}{2}=\frac{8x-2}{2}=4x-1. \]


Step 2: Solve for \(x\).
\[ 3x+2=4x-1 \ \Rightarrow\ x=3. \]
Quick Tip: For three consecutive A.P. terms \(A,B,C\), the relation \(2B=A+C\) (or \(B=\frac{A+C}{2}\)) is the quickest route.

Step 1: Identify \(a,d,l\).
\(a=2,\ d=4,\ l=82\).


Step 2: Use last-term formula.
\(l=a+(n-1)d \Rightarrow 82=2+(n-1)\cdot 4\).


Step 3: Solve for \(n\).
\(80=4(n-1)\Rightarrow n-1=20\Rightarrow n=21\).
Quick Tip: When first term \(a\), last term \(l\) and \(d\) are known: \(n=\dfrac{l-a}{d}+1\).

Step 1: Identify \(a,d\).
\(a=72,\ d=63-72=-9\).


Step 2: Set \(a_n=0\).
\(a_n=a+(n-1)d=72+(n-1)(-9)=0\).


Step 3: Solve.
\(72-9n+9=0 \Rightarrow 81-9n=0 \Rightarrow n=9\).
Quick Tip: To find when an A.P. hits a target value \(T\), solve \(a+(n-1)d=T\).

Step 1: Use the distance formula.
\[ d=\sqrt{(a\cos\theta-0)^2+(0-a\sin\theta)^2} =\sqrt{a^2\cos^2\theta+a^2\sin^2\theta}. \]


Step 2: Apply identity.
\(\cos^2\theta+\sin^2\theta=1\Rightarrow d=\sqrt{a^2}=a\) (assuming \(a\ge 0\)).
Quick Tip: Combine the Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\) with the distance formula for quick simplification.

Step 1: Use degree rule for polynomial division.

For nonzero polynomials, \(\deg(quotient)=\deg(dividend)-\deg(divisor)\) when \(\deg(dividend)\ge\deg(divisor)\).

Step 2: Apply to given polynomials.
\(\deg p=3,\ \deg q=2 \Rightarrow \deg(quotient)=3-2=1.\)
Quick Tip: While dividing polynomials, the quotient’s degree is the difference of degrees (provided leading terms don’t cancel).

Step 1: Substitute \(x=-1\).
\(g(-1)=3(-1)^2-7(-1)+2=3+7+2=12.\)
Quick Tip: Evaluate a polynomial at \(x=a\) by direct substitution; watch signs carefully.

Step 1: Factorize as difference of squares.
\(x^{2}-16=(x-4)(x+4)\).

Step 2: Set each factor to zero.
\(x=4\) or \(x=-4\).
Quick Tip: Remember \(a^{2}-b^{2}=(a-b)(a+b)\) to factor quickly.

Step 1: Use sum of roots formula.

For \(x^{2}+px+q\), \(\alpha+\beta=-p\). Here \(p=-3\Rightarrow \alpha+\beta=3\).

Step 2: Multiply by 4.
\(4(\alpha+\beta)=4\times 3=12.\)
Quick Tip: For \(ax^{2}+bx+c\): \(\alpha+\beta=-\frac{b}{a}\), \(\alpha\beta=\frac{c}{a}\).

Step 1: Use product of roots.

For \(x^{2}+px+q\), \(\alpha\beta=q\). Here \(q=2a\).

Step 2: Set equal to 8.
\(2a=8 \Rightarrow a=4.\)
Quick Tip: In a monic quadratic, the constant term equals the product of the roots.

Step 1: Factor theorem.

If \(x=a\) is a zero, then \((x-a)\) is a factor.

Step 2: Apply to \(a=-1\).

Factor is \((x-(-1))=(x+1)\).
Quick Tip: “Zero \(a\) \(\Rightarrow\) factor \((x-a)\)” is the essence of the Factor Theorem.

Step 1: Build quadratic from sum \(S\) and product \(P\).

A monic quadratic with sum \(S\) and product \(P\) is \(x^{2}-Sx+P\).

Step 2: Substitute \(S=-3,\ P=2\).
\(x^{2}-(-3)x+2 = x^{2}+3x+2\).
Quick Tip: Given sum \(S\) and product \(P\), write \(x^{2}-Sx+P\) directly.

Step 1: Use Vieta’s relation for cubics.

For \(ax^{3}+bx^{2}+cx+d\): \(\alpha\beta\gamma=-\dfrac{d}{a}\).

Step 2: State result.

Hence \(\alpha\beta\gamma=-\dfrac{d}{a}\).
Quick Tip: Signs in Vieta alternate: for cubic, product of roots is \(-d/a\).

Step 1: Degrees of factors.

First factor has degree \(2\); second has degree \(5\).

Step 2: Degree of product.
\(\deg(product)=2+5=7\).
Quick Tip: For nonzero polynomials, \(\deg(PQ)=\deg P+\deg Q\).

Step 1: Eliminate \(y\).

Multiply \(3x-2y=12\) by \(5\): \(15x-10y=60\).

Multiply \(4x-5y=16\) by \(2\): \(8x-10y=32\).

Step 2: Subtract equations.
\((15x-10y)-(8x-10y)=60-32 \Rightarrow 7x=28 \Rightarrow x=4.\)

Step 3: Back-substitute.
\(3(4)-2y=12 \Rightarrow 12-2y=12 \Rightarrow y=0.\)
Quick Tip: To solve two linear equations, align coefficients (by multiplication) and subtract to eliminate one variable.

Step 1: Formula for semicircle perimeter.

Perimeter \(= \pi r + 2r\) (half the circumference plus the diameter).

Step 2: Substitute radius \(3r\).
\[ P = \pi (3r) + 2(3r) = 3\pi r + 6r \]

Step 3: Match with options.

Oops—actual correct choice is (2) \(3\pi r + 6r\).
Quick Tip: For semicircle perimeter: add half circumference (\(\pi r\)) and diameter (\(2r\)).

Step 1: Recall probability range.

Probability always lies between 0 and 1.

Step 2: Check each option.

0.7 is valid, 75% = 0.75 valid, \(\tfrac{4}{5}=0.8\) valid.

But \(2.5 > 1\), invalid.
Quick Tip: Probability values must satisfy \(0 \leq P(E) \leq 1\).

Step 1: Complement rule.

By definition, \(E+E'=\Omega\) (the sample space).

Step 2: Apply probability axiom.
\[ P(E)+P(E')=P(\Omega)=1 \]
Quick Tip: Complementary events always add up to 1.

Step 1: Total outcomes of a die.

There are 6 equally likely outcomes.

Step 2: Favorable for not getting 2.

5 outcomes: {1,3,4,5,6}

Step 3: Probability.
\[ P(not 2)=\frac{5}{6} \]
Quick Tip: Probability = \(\frac{Favorable}{Total}\). Exclude unwanted outcome(s).

Step 1: Total cards.

52 cards in a deck.

Step 2: Black kings.

There are 2 black kings (spades, clubs).

Step 3: Probability.
\[ P=\frac{2}{52}=\frac{1}{26} \]
Quick Tip: Always count favorable cases carefully; a standard deck has 2 black suits.

Step 1: Count terms.

There are 11 numbers (odd). Median = 6th term when ordered.

Step 2: Arrange.

Sequence is already ordered: 24,27,28,31,34,\(x\),37,40,42,45,50.

Median = 6th term = \(x\).

Step 3: Use condition.

Median = 35 \(\Rightarrow x=35\).

Wait, option check: 35 corresponds to (1), not (2). Correct is (1) 35.
Quick Tip: For odd number of terms, median is the middle term directly.

Step 1: Recall usage.

Cumulative frequency helps locate median class by finding the middle observation.
Quick Tip: Cumulative frequency distribution is primarily for median calculation.

Step 1: Recall empirical relation.

Karl Pearson’s empirical formula:
\[ Mode = 3 \times Median - 2 \times Mean \]

Step 2: Rearrange.
\(Mean - Mode = 3(Mean - Median)\).
Quick Tip: The relation Mean - Mode = 3(Mean - Median) is an empirical formula, not always exact.

Step 1: Recall definition.

Modal class = class with highest frequency.

Step 2: Identify.

Frequencies: max is 45 for class 20-30.
Quick Tip: The modal class is the class interval with the maximum frequency.

Step 1: Write mean formula.
\[ \frac{(x+2)+(2x+3)+(4x+5)+(5x+2)}{4}=18 \]

Step 2: Simplify numerator.
\(x+2+2x+3+4x+5+5x+2=12x+12\).

Step 3: Equation.
\(\frac{12x+12}{4}=18 \Rightarrow 12x+12=72 \Rightarrow x=5.\)

Wait—actually value is 5, so correct answer is (1).
Quick Tip: Double-check simplification; group like terms before solving mean equations.

Step 1: Observe the segment \(PQ\).

Both \(P\) and \(Q\) have \(x\)-coordinate \(0\), so the segment \(PQ\) lies on the vertical line \(x=0\).

Step 2: Conclude for any point on the segment.

Any point \(R(a,b)\) on this segment must satisfy \(a=0\) and \(0\le b\le 2\).
Quick Tip: If two end points share the same \(x\) (or \(y\)) coordinate, the whole segment is a vertical (or horizontal) line with that coordinate fixed.

Step 1: Intersection of two lines.

Solve simultaneously: \(x=2\) and \(y=-3\).

Step 2: Read coordinates.

Point is \((2,-3)\).
Quick Tip: Lines of the form \(x=a\) and \(y=b\) intersect at \((a,b)\) immediately—no algebra needed.

Step 1: Form a right triangle.

Let opposite \(=15\), adjacent \(=8\). Then hypotenuse \(=\sqrt{15^{2}+8^{2}}=\sqrt{289}=17\).

Step 2: Compute sine.
\(\sin\theta=\dfrac{opposite}{hypotenuse}=\dfrac{15}{17}\).
Quick Tip: From \(\tan\theta=\frac{p}{b}\), set a triangle with legs \(p,b\) to get \(\sin\theta=\frac{p}{\sqrt{p^{2}+b^{2}}}\).

Step 1: Use identity.

Let \(x=\tan\theta\). Then \(x+\dfrac{1}{x}=6\).

Step 2: Square both sides.
\(\left(x+\dfrac{1}{x}\right)^{2}=x^{2}+2+\dfrac{1}{x^{2}}=36\Rightarrow x^{2}+\dfrac{1}{x^{2}}=34.\)

Thus \(\tan^{2}\theta+\cot^{2}\theta=34\).
Quick Tip: Whenever \(x+\frac{1}{x}\) is known, squaring gives \(x^{2}+\frac{1}{x^{2}}=(x+\frac{1}{x})^{2}-2\).

Step 1: Use \(\sec^{2}\theta-\tan^{2}\theta=1\).
\((\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1\).

Step 2: Substitute the given value.
\(\sec\theta+\tan\theta=-1\Rightarrow \sec\theta-\tan\theta=\dfrac{1}{-1}=-1\).
Quick Tip: Products \((\sec\pm\tan)(\sec\mp\tan)=1\) are handy consequences of \(\sec^{2}-\tan^{2}=1\).

Step 1: Compute co-functions.
\(\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\).
\(\cos\theta=\sqrt{1-\sin^{2}\theta}=\dfrac{1}{2}\Rightarrow \cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\).

Step 2: Add.
\(\csc\theta+\cot\theta=\dfrac{2\sqrt{3}}{3}+\dfrac{\sqrt{3}}{3}=\sqrt{3}\).
Quick Tip: When \(\sin\theta\) or \(\cos\theta\) is standard, compute others using \(\sin^{2}\theta+\cos^{2}\theta=1\).

Step 1: Use equality.
\(\sin\theta=\cos\theta\Rightarrow \tan\theta=1\).

Step 2: Principal value in first quadrant.
\(\theta=45^\circ\).
Quick Tip: In \(0^\circ\)–\(90^\circ\), \(\sin\theta=\cos\theta\) only at \(\theta=45^\circ\).

Step 1: Substitute standard values.
\(\sin30^\circ=\tfrac12,\ \cos45^\circ=\tfrac{\sqrt2}{2},\ \tan60^\circ=\sqrt3,\ \cot30^\circ=\sqrt3,\ \cos60^\circ=\tfrac12\).

Numerator \(= \tfrac12+\tfrac{\sqrt2}{2}-\sqrt3\).

Denominator \(= \sqrt3-\tfrac{\sqrt2}{2}-\tfrac12\).

Step 2: Multiply by 2.
\(\frac{1+\sqrt2-2\sqrt3}{\,2\sqrt3-\sqrt2-1\,}=-1\) since the numerator is the negative of the denominator.
Quick Tip: When many surds appear, clear small fractions (like halves) first—often reveals cancellations.

Step 1: Insert standard values.
\(\tan30^\circ=\tfrac{1}{\sqrt3},\ \sin30^\circ=\tfrac12,\ \cot60^\circ=\tfrac{1}{\sqrt3},\ \csc30^\circ=2\).

Step 2: Multiply.
\(\tfrac{1}{\sqrt3}\cdot \tfrac12\cdot \tfrac{1}{\sqrt3}\cdot 2=\tfrac{1}{3}\).
Quick Tip: Know the six trigonometric values at \(30^\circ,45^\circ,60^\circ\); they simplify most exam problems instantly.