CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 13

Answer: (b) tan θ = \( \frac{8}{15} \), cosec θ = \( \frac{17}{8} \), sec θ = \( \frac{17}{15} \)

Explanation:
1. For θ at X: perpendicular = YZ = 8, base = XY = 15, hypotenuse = XZ = 17.
2. tan θ = 8/15, cosec θ = hypotenuse/perpendicular = 17/8, sec θ = hypotenuse/base = 17/15.

Teacher's Note:
1. Mark the perpendicular, base and hypotenuse for the given angle before writing any ratio.
2. cosec θ uses the perpendicular and sec θ uses the base.

Answer: (b) \( \frac{1}{5} \)

Explanation:
1. \( x = \frac{\sec\theta}{5} \) and \( \frac{1}{x} = \frac{\tan\theta}{5} \).
2. \( 5\left(x^2 - \frac{1}{x^2}\right) = 5 \times \frac{\sec^2\theta - \tan^2\theta}{25} = \frac{1}{5} \), using sec²θ − tan²θ = 1.

Teacher's Note:
1. Express x and 1/x in terms of θ before squaring.
2. sec²θ − tan²θ = 1 is the key identity.

Answer: (b) \( \frac{64}{49} \)

Explanation:
1. Numerator = 1 − cos²θ = sin²θ and denominator = 1 − sin²θ = cos²θ.
2. The expression = tan²θ = \( \left(\frac{8}{7}\right)^2 = \frac{64}{49} \).

Teacher's Note:
1. Use (1 + x)(1 − x) = 1 − x² to simplify first.
2. tan θ is the reciprocal of cot θ.

Answer: (c) sec θ

Explanation:
1. \( \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} = \sqrt{\frac{(1 + \sin\theta)^2}{\cos^2\theta}} = \frac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta \).
2. \( \frac{\sin\theta(2\sin^2\theta - 1)}{\cos\theta(2\cos^2\theta - 1)} \): since 2sin²θ − 1 = −(cos²θ − sin²θ) and 2cos²θ − 1 = cos²θ − sin²θ, this equals −tan θ.
3. Sum = sec θ + tan θ − tan θ = sec θ.

Teacher's Note:
1. Multiply the numerator and the denominator of the root by (1 + sin θ).
2. 2sin²θ − 1 and 2cos²θ − 1 are opposites of each other's sign.

Answer: (a) 1

Explanation:
1. sin A = 1 − sin²A = cos²A.
2. cos²A + cos⁴A = sin A + (cos²A)² = sin A + sin²A = 1.

Teacher's Note:
1. Replace cos²A by sin A everywhere.
2. The given condition has the same form as the result.

Answer: (d) 90°

Explanation:
1. sin α = 1/2 gives α = 30° and cos β = 1/2 gives β = 60°.
2. α + β = 90°.

Teacher's Note:
1. Remember sin 30° = 1/2 and cos 60° = 1/2.
2. Both angles are acute.

Answer: (b) 1

Explanation:
1. sec²θ − 1 = tan²θ and cosec²θ − 1 = cot²θ.
2. tan²θ × cot²θ = 1.

Teacher's Note:
1. Use the identities 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ.
2. tan θ and cot θ are reciprocals.

Answer: (c) 1

Explanation:
1. sin α = √3/2 gives α = 60° and cos β = √3/2 gives β = 30°.
2. tan 60° × tan 30° = √3 × \( \frac{1}{\sqrt{3}} \) = 1.

Teacher's Note:
1. Write the angles from the standard values table.
2. tan 60° and tan 30° are reciprocals.

Answer: (a) \( \frac{1}{3} \)

Explanation:
1. Divide the numerator and denominator by cos β: \( \frac{5\tan\beta - 2}{5\tan\beta + 2} \).
2. 5 tan β = 4, so the value = \( \frac{4 - 2}{4 + 2} = \frac{1}{3} \).

Teacher's Note:
1. Divide every term by cos β to get tan β.
2. 5 tan β = 4 can be used directly.

Answer: (a) 60°

Explanation:
1. Divide the numerator and denominator on the left by cos θ: \( \frac{1 - \tan\theta}{1 + \tan\theta} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \).
2. Comparing, tan θ = √3, so θ = 60°.

Teacher's Note:
1. Compare the two sides term by term after dividing by cos θ.
2. tan 60° = √3.

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

Explanation:
1. \( \cot y° = \frac{AC}{BC} \) and \( \cot x° = \frac{AC}{CD} \).
2. \( \frac{\cot y°}{\cot x°} = \frac{CD}{BC} = \frac{1}{2} \), since D is the mid-point of BC.

Teacher's Note:
1. Use the right triangles ACD and ACB, which share the side AC.
2. D is the mid-point, so CD = BC/2.

Answer: (b) 0

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

Teacher's Note:
1. Look for a factor which is 0 before multiplying anything.
2. cos 90° = 0.

Answer: (a) \( \frac{8}{17} \)

Explanation:
1. ∠SPQ = 90°, so ∠SPT = 90° − ∠UPQ.
2. In the right triangle PUQ: cos ∠UPQ = PU/PQ = 8/17.
3. sin ∠SPT = sin(90° − ∠UPQ) = cos ∠UPQ = 8/17.

Teacher's Note:
1. The angle ∠SPT is the complement of ∠UPQ.
2. sin(90° − A) = cos A.

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

Explanation:
1. tan θ + cot θ = 2 gives tan²θ − 2 tan θ + 1 = 0, so tan θ = 1 and θ = 45°.
2. sin³45° + cos³45° = 2 × \( \left(\frac{1}{\sqrt{2}}\right)^3 \) = \( \frac{2}{2\sqrt{2}} = \frac{\sqrt{2}}{2} \).

Teacher's Note:
1. x + 1/x = 2 only for x = 1.
2. Then use the values at 45°.

Answer: (b) 2

Explanation:
1. tan α + 1/tan α = 2 gives tan α = 1.
2. So tan²⁰α + cot²⁰α = 1 + 1 = 2.

Teacher's Note:
1. The sum of a positive number and its reciprocal is 2 only when the number is 1.
2. Any power of 1 is 1.

Answer: (c) 1, 2

Explanation:
1. Divide by sin²α: cosec²α + 1 = 3 cot α, so 2 + cot²α = 3 cot α.
2. cot²α − 3 cot α + 2 = 0, so (cot α − 1)(cot α − 2) = 0 and cot α = 1 or 2.

Teacher's Note:
1. Divide both sides by sin²α to get cot α.
2. Use cosec²α = 1 + cot²α.

Answer: (a) Step 1

Explanation:
1. The identity is cosec²θ − cot²θ = 1, so 1 should be replaced by (cosec²θ − cot²θ).
2. Step 1 wrongly uses (cot²θ − cosec²θ), which equals −1.
3. Steps 2 to 4 are correct as follows from the corrected Step 1.

Teacher's Note:
1. Check the substitution for 1 first; errors usually appear in the very first step.
2. cosec²θ − cot²θ = 1, not cot²θ − cosec²θ.

Answer: (a) 0

Explanation:
1. Angles x, x, 2x sum to 180°, so x = 45° and A = B = 45°, C = 90°.
2. sec 45°/cosec 45° = √2/√2 = 1 and tan 45°/cot 45° = 1.
3. The value = 1 − 1 = 0.

Teacher's Note:
1. Find the angles first using the angle sum property.
2. At 45° many ratios are equal.

Answer: (a) −1

Explanation:
1. (sec θ + tan θ)(sec θ − tan θ) = sec²θ − tan²θ = 1.
2. sec θ + tan θ = −1, so −1 × (sec θ − tan θ) = 1 and sec θ − tan θ = −1.

Teacher's Note:
1. Use the identity sec²θ − tan²θ = 1.
2. Read the given condition as sec θ + tan θ = −1.

Answer: (c) 0

Explanation:
1. cot θ = 5/4. Divide the numerator and denominator by sin θ: \( \frac{5 - 4\cot\theta}{5 + 4\cot\theta} \).
2. 4 cot θ = 5, so the value = \( \frac{5 - 5}{5 + 5} = 0 \).

Teacher's Note:
1. Divide by sin θ to bring cot θ in.
2. 4 cot θ = 5 can be substituted directly.

Answer: (d) cos A

Explanation:
1. \( \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 - \sin A) = \frac{(1 + \sin A)(1 - \sin A)}{\cos A} \).
2. \( = \frac{1 - \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A \).

Teacher's Note:
1. Convert to sin and cos and combine.
2. (1 + sin A)(1 − sin A) = cos²A.

Answer: (a) \( \frac{2}{\sqrt{3}} \)

Explanation:
1. cos(A + 2B) = 0 gives A + 2B = 90°; cos(B − A) = √3/2 gives B − A = 30°.
2. Subtract the second equation from the first: (A + 2B) − (B − A) = 2A + B = 90° − 30° = 60°.
3. cosec 60° = \( \frac{2}{\sqrt{3}} \).

Teacher's Note:
1. Convert both conditions into angles using the standard values.
2. The combination 2A + B comes from subtracting the second from the first.

Answer: (c) 3

Explanation:
1. cosec 70° = sec 20°, so tan 20°/sec 20° = sin 20°. sec 70° = cosec 20°, so cot 20°/cosec 20° = cos 20°.
2. The first two terms give sin²20° + cos²20° = 1.
3. tan 75° tan 15° = tan 75° cot 75° = 1 and tan 45° = 1, so the last term is 2.
4. Total = 1 + 2 = 3.

Teacher's Note:
1. Convert the ratios of 70° into those of 20° using complementary angles.
2. tan(90° − θ) = cot θ.

Answer: (a) \( \frac{67}{12} \)

Explanation:
1. Numerator = \( 5 \times \frac{1}{4} + 4 \times \frac{4}{3} - 1 = \frac{5}{4} + \frac{16}{3} - 1 = \frac{15 + 64 - 12}{12} = \frac{67}{12} \).
2. Denominator = sin²30° + cos²30° = 1.
3. Value = 67/12.

Teacher's Note:
1. Use cos 60° = 1/2, sec 30° = 2/√3 and tan 45° = 1.
2. Check that the denominator is 1.

Answer: (a) a²b²

Explanation:
1. b²x² + a²y² = b²a² sin²θ + a²b² cos²θ.
2. = a²b²(sin²θ + cos²θ) = a²b².

Teacher's Note:
1. Substitute and take a²b² common.
2. sin²θ + cos²θ = 1.

Answer: (b) 2

Explanation:
1. sin x + 1/sin x = 2 holds only when sin x = 1.
2. Then cosec x = 1, so sin¹⁹x + cosec²⁰x = 1 + 1 = 2.

Teacher's Note:
1. A number plus its reciprocal equals 2 only for 1.
2. Powers of 1 are all 1.

Answer: (b) 15

Explanation:
1. sin²60° + cos²60° = 1.
2. tan 60° = √3 and tan 30° = 1/√3, so the value = \( \frac{5\sqrt{3}}{1/\sqrt{3}} = 5 \times 3 = 15 \).

Teacher's Note:
1. The bracket in the denominator is 1.
2. √3 divided by 1/√3 gives 3.

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

Explanation:
1. sin θ = cos θ gives tan θ = 1, so θ = 45°.
2. sin⁴45° + cos⁴45° = \( \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \).

Teacher's Note:
1. Equal sine and cosine mean 45°.
2. (1/√2)⁴ = 1/4.

Answer: (a) A = 45°, B = 15°

Explanation:
1. tan (A + B) = √3 gives A + B = 60° and tan (A − B) = 1/√3 gives A − B = 30°.
2. Adding: 2A = 90°, so A = 45°. Then B = 60° − 45° = 15°.

Teacher's Note:
1. Write each condition as a pair of angles.
2. Add and subtract the two equations.

Answer: (a) 9

Explanation:
1. sec²45° = 2, cosec²45° = 2 and sin 45° cos 45° = 1/2.
2. Value = 2 × 2 + 3 × 2 − 2 × \( \frac{1}{2} \) = 4 + 6 − 1 = 9.

Teacher's Note:
1. Substitute θ = 45° into every term.
2. Do not forget the factor 2 in the last term.

Answer: (a) \( \frac{119}{169} \)

Explanation:
1. tan A = 5/12, so tan²A = 25/144.
2. \( \frac{1 - 25/144}{1 + 25/144} = \frac{119/144}{169/144} = \frac{119}{169} \).

Teacher's Note:
1. Convert cot A to tan A first.
2. Use the same denominator 144.

Answer: (a) 1

Explanation:
1. cosec (A + B) = 2/√3 gives A + B = 60° and cosec (A − B) = 2 gives A − B = 30°, so A = 45°, B = 15°.
2. tan A cot A = 1 and tan B cot B = 1, so the numerator is 2. Also cosec²B cos²B = cot²B.
3. The root = \( \sqrt{\frac{2}{\cot^2 B} - \tan^2 B} = \sqrt{2\tan^2 B - \tan^2 B} = \tan B \).
4. The value = cot B × tan B = 1.

Teacher's Note:
1. Simplify the expression using identities before finding any angles.
2. cosec²B cos²B = cot²B.

Answer: (a) \( 2\left(\frac{1 + \cos^2\theta}{1 - \cos^2\theta}\right) \)

Explanation:
1. (a + b)² + (a − b)² = 2(a² + b²), with a = cot θ and b = 1/sin θ.
2. = \( 2\left(\frac{\cos^2\theta}{\sin^2\theta} + \frac{1}{\sin^2\theta}\right) = 2\,\frac{1 + \cos^2\theta}{\sin^2\theta} = 2\,\frac{1 + \cos^2\theta}{1 - \cos^2\theta} \).

Teacher's Note:
1. The cross terms cancel in (a + b)² + (a − b)².
2. Write sin²θ as 1 − cos²θ.

Answer: (b) 1

Explanation:
1. Let a sin φ = b cos φ = k. Then a = k/sin φ and b = k/cos φ.
2. a sin³φ + b cos³φ = k sin²φ + k cos²φ = k, so k = sin φ cos φ.
3. a² + b² = k²\( \left(\frac{1}{\sin^2\varphi} + \frac{1}{\cos^2\varphi}\right) \) = \( \frac{k^2}{\sin^2\varphi\cos^2\varphi} \) = 1.

Teacher's Note:
1. Introduce k for the common value a sin φ = b cos φ.
2. 1/sin² + 1/cos² = 1/(sin² cos²).

Answer: (c) A is true, but R is false.

Explanation:
1. cos⁴A − sin⁴A = (cos²A − sin²A)(cos²A + sin²A) = cos²A − sin²A = 2cos²A − 1, so A is true.
2. The correct identity is sin²θ + cos²θ = 1, so 1 + cos²θ = sin²θ is false.

Teacher's Note:
1. Factorise a² − b² for the assertion.
2. Compare R with sin²θ + cos²θ = 1.

Answer: (d) A is false, but R is true.

Explanation:
1. Divide by cos θ: \( \frac{\tan\theta + 1}{\tan\theta - 1} = \frac{4/3 + 1}{4/3 - 1} = \frac{7/3}{1/3} = 7 \), not 8, so A is false.
2. R is the Pythagoras theorem and is true.

Teacher's Note:
1. Evaluate A yourself before accepting the stated value.
2. R is always true for a right triangle.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. sin θ = cos(90° − θ), so cos(θ − 6°) = cos(90° − θ) and θ − 6° = 90° − θ.
2. 2θ = 96°, so θ = 48° and A is true; R is the identity used, so R explains A.

Teacher's Note:
1. Convert the sine into a cosine using R.
2. Then equate the two angles.

Answer: (d) A is false, but R is true.

Explanation:
1. 1/sec θ + 1/cosec θ = cos θ + sin θ, whose maximum value is √2 (at θ = 45°), not 1. So A is false.
2. The maximum of sin θ and of cos θ is each 1, so R is true.

Teacher's Note:
1. Rewrite the expression as sin θ + cos θ.
2. Both maxima occur at different angles, so they cannot be added.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. cot 12° tan 12° = 1, cot 45° = 1, tan 45° = 1 and cot 60° = 1/√3, so the product is 1/√3 and A is true.
2. R is a true identity but it is not used to get this value, so it does not explain A.

Teacher's Note:
1. Pair cot 12° with tan 12°.
2. A true reason that is not used is not an explanation.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. sin⁶α + cos⁶α = (sin²α + cos²α)³ − 3sin²α cos²α (sin²α + cos²α) = 1 − 3sin²α cos²α.
2. Adding 3sin²α cos²α gives 1, so A is true; R is the identity used to get it.

Teacher's Note:
1. Use a³ + b³ = (a + b)³ − 3ab(a + b).
2. The result comes directly from sin²α + cos²α = 1.