CBSE Class 11 Mathematics Trigonometric Functions Competitive Exam MCQs Set 01

Answer: (b) \( \frac{427\pi}{7200} \)

Explanation:
1. 10° 40′ 30″ = 10° + (40/60)° + (30/3600)° = 10.675°
2. 10.675° = \( \frac{10675}{1000} \times \frac{\pi}{180} = \frac{427\pi}{7200} \) radian

Teacher's Note:
1. Convert minutes and seconds into decimal degrees first: 1′ = (1/60)° and 1″ = (1/3600)°.
2. Then multiply by π/180 to change degrees into radians.

Answer: (d) \( \frac{5\pi}{9} \)

Explanation:
1. At 7:20 the hour hand is at 7 × 30° + 20 × 0.5° = 220° from 12.
2. The minute hand is at 20 × 6° = 120° from 12.
3. Angle between them = 220° − 120° = 100° = \( 100 \times \frac{\pi}{180} = \frac{5\pi}{9} \)

Teacher's Note:
1. The hour hand moves 0.5° every minute, so do not leave it exactly on 7.
2. The minute hand moves 6° per minute.

Answer: (d) 25 sq. cm

Explanation:
1. Area of circle = πr² = 25π, so r = 5 cm.
2. Perimeter of sector = 2r + rθ = 10 + 5θ = 20, so θ = 2 radian.
3. Area of sector = \( \frac{1}{2}r^2\theta = \frac{1}{2} \times 25 \times 2 = 25 \) sq. cm

Teacher's Note:
1. Arc length l = rθ holds only when θ is in radians.
2. The perimeter of a sector includes the two radii as well as the arc.

Answer: (c) 8

Explanation:
1. Interior angle = 3π/4 = 135°, so each exterior angle = 180° − 135° = 45°.
2. Number of sides = 360°/45° = 8.

Teacher's Note:
1. The exterior angles of any convex polygon add up to 360°.
2. Changing radians to degrees first makes such questions quick.

Answer: (a) 480°, 420°

Explanation:
1. 5π radian = 5 × 180° = 900°.
2. x + y = 900° and x − y = 60° give x = 480°, y = 420°.

Teacher's Note:
1. π radian = 180°, so 5π radian = 900°.
2. Always bring both data to the same unit before adding or subtracting.

Answer: (a) \( \frac{1}{2} \)

Explanation:
1. cos 175° = cos (180° − 5°) = −cos 5°, cos 204° = cos (180° + 24°) = −cos 24°.
2. cos 300° = cos (360° − 60°) = cos 60° = 1/2.
3. Sum = cos 24° + cos 5° − cos 5° − cos 24° + 1/2 = 1/2

Teacher's Note:
1. Look for pairs of angles that differ by 180°; their cosines cancel.
2. cos (360° − θ) = cos θ because cosine is positive in the fourth quadrant.

Answer: (b) \( \lambda \)

Explanation:
1. \( \frac{1 + \sin\alpha - \cos\alpha}{1 + \sin\alpha} = \frac{(1 + \sin\alpha - \cos\alpha)(1 + \sin\alpha + \cos\alpha)}{(1 + \sin\alpha)(1 + \sin\alpha + \cos\alpha)} \)
2. Numerator = (1 + sin α)² − cos²α = 1 + 2 sin α + sin²α − cos²α = 2 sin α + 2 sin²α = 2 sin α (1 + sin α).
3. So the expression = \( \frac{2\sin\alpha}{1 + \sin\alpha + \cos\alpha} = \lambda \)

Teacher's Note:
1. Multiplying by the 'conjugate' (1 + sin α + cos α) creates a difference of squares.
2. Replace cos²α by 1 − sin²α to make the numerator factorise.

Answer: (c) \( R - \left\{(2n + 1)\frac{\pi}{2} : n \in I\right\} \)

Explanation:
1. tan x = sin x / cos x is not defined where cos x = 0.
2. cos x = 0 when x = (2n + 1)π/2, n ∈ I.
3. So tan x is defined on R − {(2n + 1)π/2 : n ∈ I}.

Teacher's Note:
1. tan and sec share the same domain because both have cos x in the denominator.
2. cot and cosec are undefined at multiples of π, where sin x = 0.

Answer: (b) R − {nπ : n ∈ I}

Explanation:
1. cot x = cos x / sin x is not defined where sin x = 0.
2. sin x = 0 when x = nπ, n ∈ I.
3. So cot x is defined on R − {nπ : n ∈ I}.

Teacher's Note:
1. The zeros of sin x are exactly the integral multiples of π.
2. cot x is zero (not undefined) at odd multiples of π/2.

Answer: (c) \( \sec\theta = \frac{1}{2} \)

Explanation:
1. sin θ and cos θ take values in [−1, 1], so −1/5 and 1 are possible.
2. tan θ takes every real value, so 20 is possible.
3. |sec θ| ≥ 1, so sec θ = 1/2 is impossible.

Teacher's Note:
1. sec θ and cosec θ can never lie strictly between −1 and 1.
2. tan θ and cot θ have range R.

Answer: (d) a² − b² + 2ac = 0

Explanation:
1. sin θ + cos θ = b/a and sin θ cos θ = c/a.
2. (sin θ + cos θ)² = 1 + 2 sin θ cos θ ⇒ b²/a² = 1 + 2c/a.
3. b² = a² + 2ac ⇒ a² − b² + 2ac = 0

Teacher's Note:
1. Squaring sin θ + cos θ brings in sin θ cos θ, linking the sum and product of roots.
2. Multiply through by a² at the end to clear the fractions.

Answer: (d) \( \sin\theta = x + \frac{1}{x} \)

Explanation:
1. For real x ≠ 0, |x + 1/x| ≥ 2, with equality at x = ±1.
2. So sin θ = x + 1/x would need |sin θ| ≥ 2, which is impossible.
3. Options (a), (b) are possible at x = ±1, and x − 1/x takes every real value, so (c) is possible.

Teacher's Note:
1. x + 1/x never lies strictly between −2 and 2 for real x.
2. x − 1/x can be any real number, so it gives no restriction.

Answer: (a) acute angled

Explanation:
1. A triangle can have at most one obtuse angle.
2. If one angle were obtuse its cot would be negative and the other two positive, making the product negative.
3. A right angle gives a zero product. So all cots are positive: the triangle is acute angled.

Teacher's Note:
1. cot is positive in the first quadrant and negative in the second.
2. Counting how many factors can be negative is a quick way to fix the sign of a product.

Answer: (b) sin 1° < sin 1

Explanation:
1. 1 radian ≈ 57.3°, so sin 1 = sin 57.3° (approximately).
2. sin x increases on [0°, 90°], so sin 1° < sin 57.3° = sin 1.

Teacher's Note:
1. An angle written without the degree sign is in radians.
2. On [0, π/2] sine is increasing, so a bigger angle has a bigger sine.

Answer: (a) x = 45°, y = 15°

Explanation:
1. Check (a): x − y = 30°, sin 30° = 1/2; x + y = 60°, cos 60° = 1/2. Both hold.
2. Check (b): x − y = −90°, sin (−90°) = −1. Fails.
3. Check (c): x + y = 180°, cos 180° = −1. Fails.

Teacher's Note:
1. Substituting the options is the fastest method for such questions.
2. Keep the signs of negative angles in mind: sin (−θ) = −sin θ.

Answer: (b) 4

Explanation:
1. \( \frac{\sec\theta + \tan\theta}{\sec\theta - \tan\theta} = \frac{1 + \sin\theta}{1 - \sin\theta} \) (multiply up and down by cos θ)
2. With sin θ = 3/5: \( \frac{1 + \frac{3}{5}}{1 - \frac{3}{5}} = \frac{8/5}{2/5} = 4 \)

Teacher's Note:
1. Writing sec and tan in terms of sin and cos removes the need to find cos θ.
2. The answer does not depend on the sign of cos θ.

Answer: (d) \( \frac{\pi}{3} \)

Explanation:
1. Divide by 2: \( \frac{1}{2}\cos\theta + \frac{\sqrt{3}}{2}\sin\theta = 1 \)
2. i.e. cos (θ − π/3) = 1, which holds for θ = π/3.
3. Check: cos (π/3) + √3 sin (π/3) = 1/2 + 3/2 = 2.

Teacher's Note:
1. Dividing a cos θ + b sin θ by √(a² + b²) turns it into a single cosine.
2. Always verify the chosen option by direct substitution.

Answer: (a) First Quadrant; (d) Fourth Quadrant

Explanation:
1. \( \sqrt{\frac{1 - \sin A}{1 + \sin A}} = \sqrt{\frac{(1 - \sin A)^2}{\cos^2 A}} = \frac{1 - \sin A}{|\cos A|} \)
2. The equation becomes (1 − sin A)/|cos A| = (1 − sin A)/cos A, so |cos A| = cos A, i.e. cos A > 0.
3. cos A > 0 in the first and fourth quadrants.

Teacher's Note:
1. √(x²) = |x|, not x; this is where the sign condition comes from.
2. Cosine is positive in quadrants I and IV.

Answer: (d) \( \sqrt{34 - a^2} \) or \( -\sqrt{34 - a^2} \)

Explanation:
1. Let 5 sin θ + 3 cos θ = b.
2. a² + b² = (9 + 25) sin²θ + (25 + 9) cos²θ + (−30 + 30) sin θ cos θ = 34
3. b = ±√(34 − a²)

Teacher's Note:
1. Squaring and adding makes the cross terms cancel because the coefficients are arranged crosswise.
2. After taking a square root, keep both signs unless the angle's quadrant is known.

Answer: (b) 1

Explanation:
1. \( \frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8} \), so \( \tan\frac{3\pi}{8} = \cot\frac{\pi}{8} \)
2. Product = tan (π/8) cot (π/8) = 1

Teacher's Note:
1. Angles adding to π/2 have tan of one equal to cot of the other.
2. tan θ · cot θ = 1 wherever both are defined.

Answer: (b) 0

Explanation:
1. The product contains the factor cos 90°.
2. cos 90° = 0, so the whole product is 0.

Teacher's Note:
1. Before multiplying long products, look for a factor that is zero.
2. cos 90° = 0 and sin 0° = 0 are the usual culprits.

Answer: (a) 0

Explanation:
1. sin x = 1 − sin²x = cos²x.
2. cos¹²x + 3 cos¹⁰x + 3 cos⁸x + cos⁶x = (cos⁴x + cos²x)³
3. = (sin²x + sin x)³ = 1³ = 1, so the expression = 1 − 1 = 0

Teacher's Note:
1. Recognise the binomial pattern a³ + 3a²b + 3ab² + b³ = (a + b)³ with a = cos⁴x, b = cos²x.
2. The given condition says sin x = cos²x; use it to convert powers of cos into sin.

Answer: (b) \( -\frac{85}{36} \)

Explanation:
1. cos A = 8/17, sin A = 15/17; sin B = 4/5, cos B = 3/5.
2. cos (A + B) = cos A cos B − sin A sin B = 24/85 − 60/85 = −36/85
3. sec (A + B) = −85/36

Teacher's Note:
1. Use a right triangle (8, 15, 17) and (3, 4, 5) to find the missing ratios quickly.
2. A + B can exceed 90°, so a negative cosine is not surprising.

Answer: (c) \( \sqrt{2} \)

Explanation:
1. sin (3π/2 − θ) = −cos θ and cos (3π/2 − θ) = −sin θ.
2. Legs: sin θ − cos θ and cos θ + sin θ.
3. Hypotenuse² = (sin θ − cos θ)² + (sin θ + cos θ)² = 2, so hypotenuse = √2

Teacher's Note:
1. (a − b)² + (a + b)² = 2(a² + b²) is a handy shortcut.
2. In 3π/2 ± θ, sin changes to cos (and vice versa) and the sign follows the quadrant.

Answer: (a) \( \frac{1 - 2\sin^2\alpha}{2\cot\left(\frac{\pi}{4} + \alpha\right)\cos^2\left(\frac{\pi}{4} - \alpha\right)} \); (b) \( \frac{\sin(\pi - \alpha)}{\sin\alpha - \cos\alpha\tan\frac{\alpha}{2}} + \cos(\pi - \alpha) \); (c) \( \frac{1}{4\sin^2\alpha\cos^2\alpha} - \frac{(1 - \tan^2\alpha)^2}{4\tan^2\alpha} \); (d) \( \frac{1 + \sin 2\alpha}{(\sin\alpha + \cos\alpha)^2} \)

Explanation:
1. (a) cot (π/4 + α) = tan (π/4 − α), so denominator = 2 sin (π/4 − α) cos (π/4 − α) = sin (π/2 − 2α) = cos 2α = numerator. Value 1.
2. (b) sin α − cos α tan (α/2) = sin (α/2)/cos (α/2) = tan (α/2), so first term = sin α/tan (α/2) = 2cos²(α/2) = 1 + cos α; adding −cos α gives 1.
3. (c) \( \frac{1}{\sin^2 2\alpha} - \cot^2 2\alpha = \text{cosec}^2 2\alpha - \cot^2 2\alpha = 1 \), using \( \frac{1 - \tan^2\alpha}{2\tan\alpha} = \cot 2\alpha \)
4. (d) (sin α + cos α)² = 1 + sin 2α, so the value is 1. All four reduce to unity.

Teacher's Note:
1. Look for double-angle forms such as 2 sin x cos x and 1 − tan²x over 2 tan x.
2. cot (π/4 + α) = tan (π/4 − α) because the two angles add to π/2.

Answer: (b) \( \frac{255}{1313} \); (c) \( \frac{735}{1313} \)

Explanation:
1. sin A = 5/13 gives cos A = ±12/13; sin B = 99/101 gives cos B = ±20/101.
2. cos C = −cos (A + B) = sin A sin B − cos A cos B = 495/1313 − cos A cos B.
3. Both acute: cos C = (495 − 240)/1313 = 255/1313. B obtuse (B ≈ 101.4°, A ≈ 22.6°, sum < 180°): cos C = (495 + 240)/1313 = 735/1313.
4. A obtuse (≈ 157.4°) with B ≈ 78.6° exceeds 180°, so it is rejected. Hence 255/1313 or 735/1313.

Teacher's Note:
1. A given sine allows an acute or an obtuse angle; test each case against the angle sum.
2. In a triangle C = π − (A + B), so cos C = −cos (A + B).

Answer: (b) \( \sqrt{3}x^2 - (1 - a)x + a\sqrt{3} = 0 \)

Explanation:
1. tan (x + y) = \( \frac{\tan x + \tan y}{1 - a} = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} \)
2. So tan x + tan y = (1 − a)/√3 and tan x tan y = a.
3. Equation: t² − ((1 − a)/√3) t + a = 0, i.e. √3t² − (1 − a)t + a√3 = 0

Teacher's Note:
1. A quadratic with given roots is t² − (sum)t + product = 0.
2. The compound-angle formula for tan connects the sum and the product of the two tans.

Answer: (a) tan 5x − tan 3x − tan 2x

Explanation:
1. 5x = 3x + 2x, so tan 5x = \( \frac{\tan 3x + \tan 2x}{1 - \tan 3x\tan 2x} \)
2. tan 5x − tan 5x tan 3x tan 2x = tan 3x + tan 2x
3. tan 5x tan 3x tan 2x = tan 5x − tan 3x − tan 2x

Teacher's Note:
1. Whenever A = B + C, tan A tan B tan C = tan A − tan B − tan C.
2. Cross-multiplying the tan addition formula is the key move.

Answer: (a) \( \frac{\pi}{4} \)

Explanation:
1. tan (α + β) = \( \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \frac{m}{(m + 1)(2m + 1)}} = \frac{m(2m + 1) + (m + 1)}{(m + 1)(2m + 1) - m} \)
2. = (2m² + 2m + 1)/(2m² + 2m + 1) = 1
3. Hence α + β = π/4

Teacher's Note:
1. Clear the inner fractions by multiplying numerator and denominator by (m + 1)(2m + 1).
2. tan (α + β) = 1 with α, β in the standard range gives α + β = π/4.

Answer: (d) 2 sin θ

Explanation:
1. 2 + 2 cos 4θ = 4 cos²2θ, so √(2 + 2 cos 4θ) = 2|cos 2θ| = −2 cos 2θ (as 2θ is in the third quadrant).
2. 2 − 2 cos 2θ = 4 sin²θ, so the expression = 2|sin θ|.
3. π/2 < θ < 3π/4, so sin θ > 0 and the value is 2 sin θ.

Teacher's Note:
1. √(x²) = |x|; decide the sign from the quadrant of the angle.
2. 1 + cos 2x = 2 cos²x and 1 − cos 2x = 2 sin²x remove square roots.

Answer: (b) 2

Explanation:
1. tan (A − B) = 1 ⇒ (tan A − tan B)/(1 + tan A tan B) = 1
2. tan A − tan B − tan A tan B = 1
3. (1 + tan A)(1 − tan B) = 1 + tan A − tan B − tan A tan B = 1 + 1 = 2

Teacher's Note:
1. Expand the product first and compare it with the condition.
2. Similarly A + B = π/4 gives (1 + tan A)(1 + tan B) = 2.

Answer: (a) \( -\frac{627}{725} \)

Explanation:
1. Sum = 26/8 = 13/4, product = 15/8.
2. tan ((α + β)/2) = (13/4)/(1 − 15/8) = (13/4)/(−7/8) = −26/7
3. cos (α + β) = \( \frac{1 - \frac{676}{49}}{1 + \frac{676}{49}} = \frac{49 - 676}{49 + 676} = -\frac{627}{725} \)

Teacher's Note:
1. cos 2x = (1 − tan²x)/(1 + tan²x) converts a tan value straight into a cosine.
2. Here x = (α + β)/2, whose tan comes from the sum and product of the roots.

Answer: (b) \( -\frac{b}{a} \)

Explanation:
1. a = 2 sin ((α + β)/2) cos ((α − β)/2)
2. b = −2 sin ((α + β)/2) sin ((α − β)/2)
3. b/a = −tan ((α − β)/2), so tan ((α − β)/2) = −b/a

Teacher's Note:
1. Convert sums of sines and cosines into products so that common factors cancel.
2. Remember the minus sign in cos C − cos D = −2 sin ((C + D)/2) sin ((C − D)/2).

Answer: (a) 1

Explanation:
1. α + β and α − β lie in [0, π/2], so sin (α + β) = 4/5 and sin (α − β) = 3/5.
2. sin 2α = sin ((α + β) + (α − β)) = (4/5)(4/5) + (3/5)(3/5) = 16/25 + 9/25 = 1

Teacher's Note:
1. Write 2α = (α + β) + (α − β) to use the given data directly.
2. The range of the angles fixes the signs of the sines as positive.

Answer: (b) tan θ

Explanation:
1. 2 cot 2θ = (1 − tan²θ)/tan θ = cot θ − tan θ.
2. cot θ − 2 cot 2θ = cot θ − (cot θ − tan θ) = tan θ

Teacher's Note:
1. cot θ − tan θ = 2 cot 2θ is a very useful identity.
2. It is also the key step in the identity tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A.

Answer: (b) \( p\sqrt{2 - p^2} \)

Explanation:
1. (cos 20° + sin 20°)² = 2 − (cos 20° − sin 20°)² = 2 − p², so cos 20° + sin 20° = √(2 − p²) (positive).
2. cos 40° = cos²20° − sin²20° = (cos 20° − sin 20°)(cos 20° + sin 20°) = p√(2 − p²)

Teacher's Note:
1. (c + s)² + (c − s)² = 2 links the sum and difference of sine and cosine.
2. Both factors are positive here, so no negative sign is needed.

Answer: (a) 1

Explanation:
1. 2cos²x − 1 + 2 cos x = 1 ⇒ cos²x + cos x − 1 = 0 ⇒ cos²x = 1 − cos x.
2. sin²x = 1 − cos²x = cos x and 2 − cos²x = 1 + cos x.
3. sin²x (2 − cos²x) = cos x (1 + cos x) = cos x + cos²x = 1

Teacher's Note:
1. Use the relation cos²x = 1 − cos x repeatedly instead of finding cos x.
2. Expressions often collapse when written in terms of the same function.

Answer: (b) tan B − tan C − tan A

Explanation:
1. tan B = tan (A + C) = (tan A + tan C)/(1 − tan A tan C)
2. tan B − tan A tan B tan C = tan A + tan C
3. tan A tan B tan C = tan B − tan A − tan C

Teacher's Note:
1. This is the same pattern as tan 5x tan 3x tan 2x = tan 5x − tan 3x − tan 2x.
2. The angle that is the sum (here B) carries the positive sign.

Answer: (b) 2 cos 3θ

Explanation:
1. \( x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \)
2. = 8 cos³θ − 6 cos θ = 2 (4 cos³θ − 3 cos θ) = 2 cos 3θ

Teacher's Note:
1. a³ + b³ = (a + b)³ − 3ab(a + b), and here ab = 1.
2. Spot 4 cos³θ − 3 cos θ as cos 3θ.

Answer: (b) tan (α + β) = p/(q − 1)

Explanation:
1. tan α + tan β = −p and tan α tan β = q.
2. tan (α + β) = −p/(1 − q) = p/(q − 1)

Teacher's Note:
1. Use Vieta's relations: sum of roots = −p, product = q.
2. Multiply numerator and denominator by −1 to match the option.

Answer: (b) \( \frac{1}{8} \)

Explanation:
1. The product is sin 10° sin 50° sin 70° = sin 10° sin (60° − 10°) sin (60° + 10°).
2. sin θ sin (60° − θ) sin (60° + θ) = (1/4) sin 3θ, so the product = (1/4) sin 30° = 1/8

Teacher's Note:
1. sin (60° − θ) sin (60° + θ) = sin²60° − sin²θ = 3/4 − sin²θ; multiplying by sin θ gives (1/4) sin 3θ.
2. Write the angles in degrees to spot the 60° ± θ pattern.

Answer: (a) \( \frac{1}{64} \)

Explanation:
1. Use cos θ cos (60° − θ) cos (60° + θ) = (1/4) cos 3θ.
2. θ = 12°: cos 12° cos 48° cos 72° = (1/4) cos 36°; θ = 24°: cos 24° cos 36° cos 84° = (1/4) cos 72°.
3. Product = (1/16) cos 36° cos 72° = (1/16) × (√5 + 1)/4 × (√5 − 1)/4 = (1/16) × (1/4) = 1/64

Teacher's Note:
1. Grouping angles as θ, 60° − θ, 60° + θ turns a long product into short ones.
2. cos 36° cos 72° = 1/4 follows from cos 36° = (√5 + 1)/4 and cos 72° = sin 18° = (√5 − 1)/4.

Answer: (b) \( -\frac{1}{2} \)

Explanation:
1. (sin 78° − sin 42°) − (sin 66° − sin 6°) = 2 cos 60° sin 18° − 2 cos 36° sin 30°
2. = sin 18° − cos 36° = (√5 − 1)/4 − (√5 + 1)/4
3. = −2/4 = −1/2

Teacher's Note:
1. Pair the terms so that the half-sum becomes a standard angle (60° and 36°).
2. Know sin 18° = (√5 − 1)/4 and cos 36° = (√5 + 1)/4.