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Mathematics - MCQ Practice Questions

Mathematics questions across competitive exams reuse a small set of ideas in many disguises, so the aim is recognition rather than memorisation. This set covers algebra, geometry and mensuration, trigonometry, number theory, statistics and probability, and applied arithmetic. Every solution shows the full working, and where a shorter route exists it is shown alongside so you can judge which one suits your speed.

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All topics Vectors & 3D Sets & Functions Algebra Geometry Trigonometry Calculus Statistics & Probability Number Theory Matrices & Determinants Coordinate Geometry Sets & Relations Mathematical Reasoning

Q.1Medium

If $sin θ + cos θ = 2 cos θ$ , then $cot θ$ equals:

Answer: B

Understanding:

We are given $sin θ + cos θ = 2 cos θ$ and must find $cot θ$ .

Formula:

cot θ = \frac{cos θ}{sin θ}

Step 1: Rearrange the given equation.

sin θ + cos θ sin θ sin θ = 2 cos θ = 2 cos θ - cos θ = (2 - 1) cos θ

Step 2: Divide both sides by $sin θ$ .

1 cot θ = (2 - 1) \cdot \frac{cos θ}{sin θ} = \frac{1}{2 - 1}

Step 3: Rationalise the denominator.

cot θ = \frac{1}{2 - 1} \times \frac{2 + 1}{2 + 1} = \frac{2 + 1}{2 - 1} = 2 + 1

Answer:

The value of $cot θ$ is $2 + 1$ .

cot θ = 2 + 1

Quick Tip:

Rationalising $\frac{1}{2 - 1}$ is a very common step in trigonometry problems — always multiply by the conjugate.

Q.2Medium

The value of $sin 18°$ is:

Answer: B

Understanding:

We need the exact value of $sin 18°$ .

Formula:

Let $θ = 18°$ , so $5 θ = 90°$ , giving $2 θ = 90° - 3 θ$ .

sin 2 θ = cos 3 θ

Step 1: Expand both sides using standard identities.

2 sin θ cos θ = 4 cos^{3} θ - 3 cos θ

Step 2: Divide both sides by $cos θ$ (which is non-zero for $θ = 18°$ ).

2 sin θ = 4 cos^{2} θ - 3 = 4 (1 - sin^{2} θ) - 3 = 1 - 4 sin^{2} θ

Step 3: Rearrange into a quadratic in $sin θ$ .

4 sin^{2} θ + 2 sin θ - 1 sin θ = 0 = \frac{- 2 \pm 4 + 16}{8} = \frac{- 2 \pm 20}{8} = \frac{- 1 \pm 5}{4}

Step 4: Choose the positive root since $sin 18° > 0$ .

sin 18° = \frac{5 - 1}{4}

Answer:

The exact value of $sin 18°$ is $\frac{5 - 1}{4}$ .

sin 18° = \frac{5 - 1}{4}

Quick Tip:

$cos 36° = \frac{5 + 1}{4}$ is the companion result — both are frequently tested in competitive exams.

Q.3Medium

If $tan A = \frac{1}{2}$ and $tan B = \frac{1}{3}$ , then $A + B$ equals:

Answer: C

Understanding:

We are given $tan A = \frac{1}{2}$ and $tan B = \frac{1}{3}$ and must find $A + B$ .

•

tan A = \frac{1}{2}

•

tan B = \frac{1}{3}

Formula:

tan (A + B) = \frac{tan A + tan B}{1 - tan A \cdot tan B}

Step 1: Substitute the given values.

tan (A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}}

Step 2: Simplify numerator and denominator.

Numerator Denominator = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} = 1 - \frac{1}{6} = \frac{5}{6}

Step 3: Compute the ratio.

tan (A + B) = \frac{\frac{5}{6}}{\frac{5}{6}} = 1

Step 4: Find $A + B$ .

A + B = arctan (1) = \frac{π}{4}

Answer:

The value of $A + B$ is $\frac{π}{4}$ .

A + B = \frac{π}{4}

Quick Tip:

Whenever $tan (A + B) = 1$ with $A, B$ being small positive angles, $A + B = 45°$ is the expected principal value.

Q.4Medium

The general solution of $sin θ = - \frac{1}{2}$ is:

Answer: C

Understanding:

We need the general solution of $sin θ = - \frac{1}{2}$ .

Formula:

The general solution of $sin θ = sin α$ is:

θ = nπ + (- 1)^{n} α, n \in Z

Step 1: Identify $α$ .

Since $sin (- \frac{π}{6}) = - \frac{1}{2}$ , the principal value is $α = - \frac{π}{6}$ .

Step 2: Write the general solution.

θ = nπ + (- 1)^{n} (- \frac{π}{6}), n \in Z

Verification:

•For

n = 0

θ = - \frac{π}{6}

sin (- \frac{π}{6}) = - \frac{1}{2}

✓

•For

n = 1

θ = π + \frac{π}{6} = \frac{7 π}{6}

sin \frac{7 π}{6} = - \frac{1}{2}

✓

Answer:

The general solution is $θ = nπ + (- 1)^{n} (- \frac{π}{6})$ .

θ = nπ + (- 1)^{n} (- \frac{π}{6}), n \in Z

Quick Tip:

The general solution of $sin θ = k$ always uses the principal value $α = arcsin (k)$ , which can be negative — never force $α$ to be positive.

Q.5Medium

The value of $cos^{2} 48° - sin^{2} 12°$ is:

Answer: B

Understanding:

We must evaluate $cos^{2} 48° - sin^{2} 12°$ .

Formula:

cos^{2} A - sin^{2} B = cos (A + B) cos (A - B)

Step 1: Identify $A = 48°$ and $B = 12°$ , then apply the identity.

cos^{2} 48° - sin^{2} 12° = cos (48° + 12°) \cdot cos (48° - 12°) = cos 60° \cdot cos 36°

Step 2: Substitute exact values.

cos 60° = \frac{1}{2}, cos 36° = \frac{5 + 1}{4}

Step 3: Multiply.

cos 60° \cdot cos 36° = \frac{1}{2} \times \frac{5 + 1}{4} = \frac{5 + 1}{8}

Verification:

Numerically, $cos^{2} 48° \approx 0.4472$ and $sin^{2} 12° \approx 0.0432$ , so the difference $\approx 0.404$ . Also $\frac{5 + 1}{8} \approx \frac{3.236}{8} \approx 0.405$ ✓

Answer:

The value of $cos^{2} 48° - sin^{2} 12°$ is $\frac{5 + 1}{8}$ .

\frac{5 + 1}{8}

Quick Tip:

The identity $cos^{2} A - sin^{2} B = cos (A + B) cos (A - B)$ is the key tool for this class of problems.

Q.6Medium

If $sin θ + sin^{2} θ = 1$ , then $cos^{2} θ + cos^{4} θ$ equals:

Answer: C

Understanding:

Given $sin θ + sin^{2} θ = 1$ , we must find the value of $cos^{2} θ + cos^{4} θ$ .

Formula:

The Pythagorean identity:

sin^{2} θ + cos^{2} θ = 1 ⟹ cos^{2} θ = 1 - sin^{2} θ

Step 1: From the given equation, express $sin θ$ in terms of $cos^{2} θ$ .

sin θ + sin^{2} θ sin θ = 1 = 1 - sin^{2} θ = cos^{2} θ

Step 2: Now compute $cos^{2} θ + cos^{4} θ$ .

cos^{2} θ + cos^{4} θ = cos^{2} θ (1 + cos^{2} θ)

Step 3: Substitute $cos^{2} θ = sin θ$ .

= sin θ (1 + sin θ) = sin θ + sin^{2} θ = 1 (from the given condition)

Answer:

The value of $cos^{2} θ + cos^{4} θ$ is $1$ .

cos^{2} θ + cos^{4} θ = 1

Quick Tip:

The key insight is recognising that $sin θ = 1 - sin^{2} θ = cos^{2} θ$ — this substitution converts the target expression back to the given condition.

Q.7Medium

The maximum value of $3 sin θ + 4 cos θ$ is:

Answer: D

Understanding:

We must find the maximum value of the expression $3 sin θ + 4 cos θ$ .

Formula:

The maximum value of $a sin θ + b cos θ$ is:

Maximum = a^{2} + b^{2}

Step 1: Identify $a$ and $b$ .

•

a = 3

b = 4

Step 2: Apply the formula.

Maximum = 3^{2} + 4^{2} = 9 + 16 = 25 = 5

Verification:

Write $3 sin θ + 4 cos θ = 5 (\frac{3}{5} sin θ + \frac{4}{5} cos θ) = 5 sin (θ + ϕ)$ where $cos ϕ = \frac{3}{5}$ . The maximum of $5 sin (θ + ϕ)$ is indeed $5$ ✓

Answer:

The maximum value of $3 sin θ + 4 cos θ$ is $5$ .

5

Quick Tip:

The $3$ - $4$ - $5$ right triangle appears here: the coefficients form a Pythagorean triple, making $3^{2} + 4^{2} = 5$ immediate.

Q.8Medium

If $cos α + cos β = 0$ and $sin α + sin β = 0$ , then $cos (α - β)$ equals:

Answer: C

Understanding:

Given $cos α + cos β = 0$ and $sin α + sin β = 0$ , we must find $cos (α - β)$ .

Formula:

cos (α - β) = cos α cos β + sin α sin β

Step 1: From the given equations, write:

cos β = - cos α and sin β = - sin α

Step 2: This means $β = π + α$ (or equivalently $α$ and $β$ differ by $π$ ).

Step 3: Substitute into the formula.

cos (α - β) = cos α \cdot (- cos α) + sin α \cdot (- sin α) = - cos^{2} α - sin^{2} α = - (cos^{2} α + sin^{2} α) = - 1

Verification:

If $α = 0°$ and $β = 180°$ , then $cos 0° + cos 180° = 1 - 1 = 0$ ✓ and $sin 0° + sin 180° = 0$ ✓. Also, $cos (0° - 180°) = cos (- 180°) = - 1$ ✓

Answer:

The value of $cos (α - β)$ is $- 1$ .

cos (α - β) = - 1

Quick Tip:

The conditions $cos α = - cos β$ and $sin α = - sin β$ together force $α$ and $β$ to be supplementary in the sense $β = α \pm π$ , which always gives $cos (α - β) = - 1$ .

Q.9Medium

The value of $tan 75°$ is:

Answer: C

Understanding:

We must find the exact value of $tan 75°$ .

Formula:

tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}

Step 1: Write $75° = 45° + 30°$ .

tan 75° = tan (45° + 30°)

Step 2: Apply the addition formula with $tan 45° = 1$ and $tan 30° = \frac{1}{3}$ .

tan 75° = \frac{1 + \frac{1}{3}}{1 - 1 \cdot \frac{1}{3}} = \frac{\frac{3 + 1}{3}}{\frac{3 - 1}{3}} = \frac{3 + 1}{3 - 1}

Step 3: Rationalise.

tan 75° = \frac{( 3 + 1 ) ^{2}}{( 3 - 1 ) ( 3 + 1 )} = \frac{3 + 2 3 + 1}{3 - 1} = \frac{4 + 2 3}{2} = 2 + 3

Verification:

Numerically, $tan 75° \approx 3.732$ and $2 + 3 \approx 2 + 1.732 = 3.732$ ✓

Answer:

The value of $tan 75°$ is $2 + 3$ .

tan 75° = 2 + 3

Quick Tip:

$tan 75°$ and $tan 15° = 2 - 3$ are reciprocals of each other — a quick way to cross-check since $tan 75° \cdot tan 15° = 1$ .

Q.10Medium

If $α$ and $β$ are the solutions of $a cos θ + b sin θ = c$ , then $cos (α + β)$ equals:

Answer: A

Understanding:

$α$ and $β$ are both solutions of $a cos θ + b sin θ = c$ . We need $cos (α + β)$ .

Formula:

For $α$ and $β$ satisfying $a cos θ + b sin θ = c$ , use the Weierstrass (half-angle) substitution $t = tan \frac{θ}{2}$ :

cos θ = \frac{1 - t ^{2}}{1 + t ^{2}}, sin θ = \frac{2 t}{1 + t ^{2}}

Step 1: Substitute into the equation.

a \cdot \frac{1 - t ^{2}}{1 + t ^{2}} + b \cdot \frac{2 t}{1 + t ^{2}} = c

a (1 - t^{2}) + 2 b t = c (1 + t^{2})

(a + c) t^{2} - 2 b t + (c - a) = 0

Step 2: The roots are $t_{1} = tan \frac{α}{2}$ and $t_{2} = tan \frac{β}{2}$ . By Vieta's formulas:

t_{1} + t_{2} = \frac{2 b}{a + c}, t_{1} t_{2} = \frac{c - a}{a + c}

Step 3: Use $tan \frac{α + β}{2} = \frac{t _{1} + t _{2}}{1 - t _{1} t _{2}}$ .

tan \frac{α + β}{2} = \frac{\frac{2 b}{a + c}}{1 - \frac{c - a}{a + c}} = \frac{\frac{2 b}{a + c}}{\frac{a + c - c + a}{a + c}} = \frac{2 b}{2 a} = \frac{b}{a}

Step 4: Compute $cos (α + β)$ using $cos (α + β) = \frac{1 - tan ^{2} \frac{α + β}{2}}{1 + tan ^{2} \frac{α + β}{2}}$ .

cos (α + β) = \frac{1 - \frac{b ^{2}}{a ^{2}}}{1 + \frac{b ^{2}}{a ^{2}}} = \frac{a ^{2} - b ^{2}}{a ^{2} + b ^{2}}

Answer:

The value of $cos (α + β)$ is $\frac{a ^{2} - b ^{2}}{a ^{2} + b ^{2}}$ .

cos (α + β) = \frac{a ^{2} - b ^{2}}{a ^{2} + b ^{2}}

Quick Tip:

The Weierstrass substitution converts a trigonometric equation into a quadratic, allowing Vieta's formulas to extract symmetric functions of the roots elegantly.

Q.11Medium

If $sin θ + cos θ = 2 cos θ$ , then the value of $tan θ$ is:

Answer: A

Understanding:

We are given $sin θ + cos θ = 2 cos θ$ and must find $tan θ$ .

Formula:

tan θ = \frac{sin θ}{cos θ}

Step 1: Isolate $sin θ$ .

sin θ + cos θ sin θ sin θ = 2 cos θ = 2 cos θ - cos θ = (2 - 1) cos θ

Step 2: Divide both sides by $cos θ$ .

\frac{sin θ}{cos θ} tan θ = 2 - 1 = 2 - 1

Answer:

The value of $tan θ$ is $2 - 1$ .

tan θ = 2 - 1

Quick Tip:

A common trap is to transpose $cos θ$ to the other side incorrectly and get $2 + 1$ . Always isolate $sin θ$ first, then divide.

Q.12Medium

The value of $sin 20° \cdot sin 40° \cdot sin 80°$ is:

Answer: A

Understanding:

We need to evaluate $sin 20° \cdot sin 40° \cdot sin 80°$ .

Formula:

For any angle $θ$ , the following product identity holds:

sin θ \cdot sin (60° - θ) \cdot sin (60° + θ) = \frac{1}{4} sin 3 θ

Step 1: Recognise the pattern with $θ = 20°$ .

sin 20° \cdot sin (60° - 20°) \cdot sin (60° + 20°) sin 20° \cdot sin 40° \cdot sin 80° = \frac{1}{4} sin (3 \times 20°) = \frac{1}{4} sin 60°

Step 2: Substitute $sin 60° = \frac{3}{2}$ .

= \frac{1}{4} \times \frac{3}{2} = \frac{3}{8}

Answer:

The value of $sin 20° \cdot sin 40° \cdot sin 80°$ is $\frac{3}{8}$ .

\frac{3}{8}

Quick Tip:

The identity $sin θ \cdot sin (60° - θ) \cdot sin (60° + θ) = \frac{1}{4} sin 3 θ$ is extremely useful in competitive exams and worth memorising.

Q.13Medium

The value of $cos^{2} 10° - cos 10° cos 50° + cos^{2} 50°$ is:

Answer: A

Understanding:

We need to evaluate $cos^{2} 10° - cos 10° cos 50° + cos^{2} 50°$ .

Formula:

We use the identity $a^{2} - ab + b^{2} = (a - b)^{2} + ab$ and the product-to-sum formula:

cos A cos B = \frac{1}{2} [cos (A - B) + cos (A + B)]

Step 1: Use the substitution. Let $a = cos 10°$ , $b = cos 50°$ .

a^{2} - ab + b^{2} = (a^{2} + b^{2}) - ab

Step 2: Compute $a^{2} + b^{2} = cos^{2} 10° + cos^{2} 50°$ .

cos^{2} 10° + cos^{2} 50° = \frac{1 + cos 20°}{2} + \frac{1 + cos 100°}{2} = 1 + \frac{cos 20° + cos 100°}{2}

Step 3: Apply sum-to-product: $cos 20° + cos 100° = 2 cos 60° cos 40° = 2 \cdot \frac{1}{2} \cdot cos 40° = cos 40°$ .

cos^{2} 10° + cos^{2} 50° = 1 + \frac{cos 40°}{2}

Step 4: Compute $ab = cos 10° cos 50°$ .

cos 10° cos 50° = \frac{1}{2} [cos 40° + cos 60°] = \frac{1}{2} [cos 40° + \frac{1}{2}] = \frac{cos 40°}{2} + \frac{1}{4}

Step 5: Combine.

a^{2} - ab + b^{2} = (1 + \frac{cos 40°}{2}) - (\frac{cos 40°}{2} + \frac{1}{4}) = 1 + \frac{cos 40°}{2} - \frac{cos 40°}{2} - \frac{1}{4} = \frac{3}{4}

Answer:

The value of the expression is $\frac{3}{4}$ .

\frac{3}{4}

Quick Tip:

For expressions of the form $cos^{2} A - cos A cos (A + 40°) + cos^{2} (A + 40°)$ , the result is always $\frac{3}{4}$ — a useful pattern in competitive exams.

Q.14Medium

If $tan A = \frac{1}{2}$ and $tan B = \frac{1}{3}$ , then the value of $A + B$ (where $A, B$ are acute) is:

Answer: A

Understanding:

We are given $tan A = \frac{1}{2}$ and $tan B = \frac{1}{3}$ and need to find $A + B$ .

•

tan A = \frac{1}{2}

•

tan B = \frac{1}{3}

Formula:

tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}

Step 1: Substitute the values.

tan (A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}}

Step 2: Simplify the numerator and denominator.

Numerator Denominator = \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6} = 1 - \frac{1}{6} = \frac{5}{6}

Step 3: Compute $tan (A + B)$ .

tan (A + B) = \frac{\frac{5}{6}}{\frac{5}{6}} = 1

Step 4: Since $A$ and $B$ are acute, $A + B \in (0°, 180°)$ and $tan (A + B) = 1$ gives:

A + B = 45°

Answer:

The value of $A + B$ is $45°$ .

A + B = 45°

Q.15Medium

The general solution of $sin 2 θ = cos θ$ is:

Answer: A

Understanding:

We need the general solution of $sin 2 θ = cos θ$ .

Formula:

sin 2 θ = 2 sin θ cos θ

Step 1: Rewrite the equation.

2 sin θ cos θ 2 sin θ cos θ - cos θ cos θ (2 sin θ - 1) = cos θ = 0 = 0

Step 2: Set each factor to zero.

cos θ = 0 or sin θ = \frac{1}{2}

Step 3: Solve $cos θ = 0$ .

θ = (2 n + 1) \frac{π}{2}, n \in Z

Step 4: Solve $sin θ = \frac{1}{2}$ .

θ = nπ + (- 1)^{n} \frac{π}{6}, n \in Z

Answer:

The general solution is $θ = (2 n + 1) \frac{π}{2}$ or $θ = nπ + (- 1)^{n} \frac{π}{6}$ .

θ = (2 n + 1) \frac{π}{2} or θ = nπ + (- 1)^{n} \frac{π}{6}

Quick Tip:

Always factorise trigonometric equations rather than dividing by a trig function — dividing by $cos θ$ would lose the solutions where $cos θ = 0$ .

Q.16Medium

The value of $\frac{sin 5 A - sin 3 A}{cos 3 A + cos 5 A}$ is:

Answer: A

Understanding:

We need to simplify $\frac{sin 5 A - sin 3 A}{cos 3 A + cos 5 A}$ .

Formula:

Sum-to-product identities:

sin C - sin D = 2 cos \frac{C + D}{2} sin \frac{C - D}{2}

cos C + cos D = 2 cos \frac{C + D}{2} cos \frac{C - D}{2}

Step 1: Apply to the numerator with $C = 5 A$ , $D = 3 A$ .

sin 5 A - sin 3 A = 2 cos \frac{5 A + 3 A}{2} sin \frac{5 A - 3 A}{2} = 2 cos 4 A sin A

Step 2: Apply to the denominator with $C = 3 A$ , $D = 5 A$ .

cos 3 A + cos 5 A = 2 cos \frac{3 A + 5 A}{2} cos \frac{5 A - 3 A}{2} = 2 cos 4 A cos A

Step 3: Divide.

\frac{sin 5 A - sin 3 A}{cos 3 A + cos 5 A} = \frac{2 cos 4 A sin A}{2 cos 4 A cos A} = \frac{sin A}{cos A} = tan A

Answer:

The expression simplifies to $tan A$ .

tan A

Quick Tip:

Whenever you see a difference of sines over a sum of cosines (or vice versa) with arguments in arithmetic progression, sum-to-product identities reduce the expression to a single trig ratio in one step.

Q.17Medium

If $3 sin θ + 4 cos θ = 5$ , then the value of $3 cos θ - 4 sin θ$ is:

Answer: A

Understanding:

Given $3 sin θ + 4 cos θ = 5$ , we must find $3 cos θ - 4 sin θ$ .

Formula:

For any expression of the form $a sin θ + b cos θ$ , the maximum value is $a^{2} + b^{2}$ .

If (a sin θ + b cos θ)^{2} + (a cos θ - b sin θ)^{2} = a^{2} + b^{2}

Step 1: Square both expressions and add them.

(3 sin θ + 4 cos θ)^{2} + (3 cos θ - 4 sin θ)^{2} = 9 sin^{2} θ + 24 sin θ cos θ + 16 cos^{2} θ + 9 cos^{2} θ - 24 sin θ cos θ + 16 sin^{2} θ = 9 (sin^{2} θ + cos^{2} θ) + 16 (cos^{2} θ + sin^{2} θ) = 9 + 16 = 25

Step 2: Substitute the known value.

5^{2} + (3 cos θ - 4 sin θ)^{2} 25 + (3 cos θ - 4 sin θ)^{2} (3 cos θ - 4 sin θ)^{2} 3 cos θ - 4 sin θ = 25 = 25 = 0 = 0

Answer:

The value of $3 cos θ - 4 sin θ$ is $0$ .

0

Quick Tip:

The identity $(a sin θ + b cos θ)^{2} + (a cos θ - b sin θ)^{2} = a^{2} + b^{2}$ is a powerful tool. When the first expression attains the maximum $a^{2} + b^{2}$ , the second must equal $0$ .

Q.18Medium

The value of $cos 1° \cdot cos 2° \cdot cos 3° \dots cos 90°$ is:

Answer: A

Understanding:

We need to evaluate the product $cos 1° \cdot cos 2° \cdot cos 3° \dots cos 90°$ .

Formula:

A product is zero if any one of its factors equals zero.

If any factor = 0, then the entire product = 0

Step 1: Identify whether any factor in the product is zero.

cos 90° = 0

Step 2: Since $cos 90°$ is one of the factors in the product, the entire product equals zero regardless of the other factors.

cos 1° \cdot cos 2° \dots cos 89° \cdot cos 90° = cos 1° \cdot cos 2° \dots cos 89° \cdot 0 = 0

Answer:

The value of the product is $0$ .

0

Quick Tip:

Before computing any complex product, always check whether any single factor equals zero — that immediately makes the entire product zero.

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