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Mathematics - MCQ Practice Questions

Practice free Mathematics multiple-choice questions with detailed answers and explanations. Perfect for competitive exam preparation.

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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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