CBSE Class 11 Mathematics Trigonometric Functions Competitive Exam MCQs Set 02

Answer: (b) sin 2x = 0; (d) cos x = −1/2

Explanation:
1. sin x + sin 3x = 2 sin 2x cos x.
2. So the sum = sin 2x (2 cos x + 1).
3. It vanishes when sin 2x = 0 or cos x = −1/2.

Teacher's Note:
1. Pair the outer terms so that a common factor sin 2x appears.
2. A product is zero when any factor is zero.

Answer: (b) tan A tan 2A tan 3A

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

Teacher's Note:
1. Again the pattern A = B + C gives tan A − tan B − tan C = tan A tan B tan C.
2. This identity is useful for summing or simplifying such expressions.

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

Explanation:
1. cos (7π/8) = −cos (π/8) and cos (5π/8) = −cos (3π/8).
2. Product = cos²(π/8) cos²(3π/8) = cos²(π/8) sin²(π/8)
3. = (1/4) sin²(π/4) = (1/4)(1/2) = 1/8

Teacher's Note:
1. Angles that add to π have cosines of opposite sign.
2. cos (3π/8) = sin (π/8) because 3π/8 = π/2 − π/8.

Answer: (d) 4

Explanation:
1. 5 sin²θ + 4 cos²θ = 4 (sin²θ + cos²θ) + sin²θ = 4 + sin²θ
2. sin²θ ≥ 0, so the minimum is 4 (at θ = 0).

Teacher's Note:
1. Rewrite the expression so that only one trigonometric function remains.
2. 0 ≤ sin²θ ≤ 1 then gives the range [4, 5] at once.

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

Explanation:
1. sin x + cos x = √2 sin (x + π/4).
2. The maximum of sin is 1, so the maximum value is √2.

Teacher's Note:
1. a sin x + b cos x always lies between −√(a² + b²) and √(a² + b²).
2. Here a = b = 1, so the bound is √2.

Answer: (b) 4

Explanation:
1. \( \sec^2\theta + \text{cosec}^2\theta = \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{4}{\sin^2 2\theta} \)
2. sin²2θ ≤ 1, so the expression ≥ 4, attained at θ = π/4.

Teacher's Note:
1. sec²θ + cosec²θ = sec²θ cosec²θ, which is a neat identity.
2. Writing sin θ cos θ as (1/2) sin 2θ brings in a known range.

Answer: (d) \( \frac{1}{4} \)

Explanation:
1. sin⁶θ + cos⁶θ = (sin²θ + cos²θ)³ − 3 sin²θ cos²θ (sin²θ + cos²θ) = 1 − 3 sin²θ cos²θ
2. = 1 − (3/4) sin²2θ
3. Minimum when sin²2θ = 1: 1 − 3/4 = 1/4

Teacher's Note:
1. Use a³ + b³ = (a + b)³ − 3ab(a + b) with a = sin²θ, b = cos²θ.
2. Similarly sin⁴θ + cos⁴θ = 1 − (1/2) sin²2θ.

Answer: (a) [−1, 3]

Explanation:
1. cos 2θ = 1 − 2 sin²θ, so A = 2 sin²θ − 1 + 2 sin²θ = 4 sin²θ − 1.
2. 0 ≤ sin²θ ≤ 1 gives −1 ≤ A ≤ 3.

Teacher's Note:
1. Express everything in sin²θ, whose range is [0, 1].
2. Then apply the linear change carefully to both ends.

Answer: (d) 0

Explanation:
1. Each sine is at most 1, so the sum 3 forces sin θ₁ = sin θ₂ = sin θ₃ = 1.
2. Then each cosine = 0, so the sum is 0.

Teacher's Note:
1. When a sum of bounded quantities reaches its maximum, every term must be at its maximum.
2. sin θ = 1 means cos θ = 0, since sin²θ + cos²θ = 1.

Answer: (b) 4

Explanation:
1. 12 sin θ − 9 sin²θ = 4 − (3 sin θ − 2)²
2. The square is 0 when sin θ = 2/3 (allowed), so the maximum is 4.

Teacher's Note:
1. Complete the square in sin θ and check the vertex value lies in [−1, 1].
2. If the vertex lay outside [−1, 1], the extreme value would occur at sin θ = ±1.

Answer: (b) 0

Explanation:
1. cos θ + √3 sin θ = 2 cos (θ − π/3), which lies in [−2, 2].
2. It can never equal 5, so there is no solution.

Teacher's Note:
1. Before solving, check that the right side lies within the range of the left side.
2. Range of a cos θ + b sin θ is [−√(a² + b²), √(a² + b²)].

Answer: (c) \( \frac{11}{32} \)

Explanation:
1. \( \sin\frac{A}{2}\sin\frac{5A}{2} = \frac{1}{2}(\cos 2A - \cos 3A) \)
2. cos 2A = 2(9/16) − 1 = 1/8; cos 3A = 4(27/64) − 3(3/4) = −9/16
3. Value = (1/2)(1/8 + 9/16) = (1/2)(11/16) = 11/32

Teacher's Note:
1. Turn the product into a difference using 2 sin A sin B = cos (A − B) − cos (A + B).
2. Then use the double and triple angle formulas in terms of cos A.

Answer: (a) ≥ 0 for all θ

Explanation:
1. sin θ + sin 3θ = 2 sin 2θ cos θ.
2. sin θ (sin θ + sin 3θ) = 2 sin θ cos θ sin 2θ = sin²2θ ≥ 0 for all θ.

Teacher's Note:
1. A perfect square of a real number is never negative.
2. Sum-to-product often reveals such hidden squares.

Answer: (c) \( \sqrt{3} \)

Explanation:
1. \( \tan^2\frac{\alpha}{2} = \frac{1 - \cos\alpha}{1 + \cos\alpha} = \frac{(2 - \cos\beta) - (2\cos\beta - 1)}{(2 - \cos\beta) + (2\cos\beta - 1)} = \frac{3(1 - \cos\beta)}{1 + \cos\beta} \)
2. = 3 tan²(β/2)
3. Both half-angles lie in (0, π/2), so tan (α/2) cot (β/2) = √3

Teacher's Note:
1. tan²(x/2) = (1 − cos x)/(1 + cos x) is the natural bridge between cos x and half angles.
2. The given ranges make both tan (α/2) and tan (β/2) positive.

Answer: (b) (a² + b²) cos θ = 2ab

Explanation:
1. (a) sin²θ = 5/4 > 1: impossible.
2. (b) cos θ = 2ab/(a² + b²); since a² + b² ≥ 2|ab|, |cos θ| ≤ 1: possible.
3. (c) cosec θ = (m² − n²)/(m² + n²) has modulus less than 1 for non-zero m, n: impossible.

Teacher's Note:
1. Compare each expression with the allowed range of the function.
2. a² + b² ≥ 2|ab| follows from (|a| − |b|)² ≥ 0.

Answer: (c) 1

Explanation:
1. cos A (3 − 4cos²A) = 3 cos A − 4 cos³A = −cos 3A.
2. sin A (3 − 4 sin²A) = 3 sin A − 4 sin³A = sin 3A.
3. Expression = cos²3A + sin²3A = 1

Teacher's Note:
1. Push one factor of cos A or sin A inside the bracket to recognise triple-angle forms.
2. Squares remove the minus sign in −cos 3A.

Answer: (c) cos 2α

Explanation:
1. 4 sin α sin β cos (α + β) = 2 cos (α + β)[cos (α − β) − cos (α + β)] = 2 cos (α + β) cos (α − β) − 2cos²(α + β)
2. = (cos 2α + cos 2β) − 2cos²(α + β); also cos 2(α + β) = 2cos²(α + β) − 1 and 2 sin²β = 1 − cos 2β
3. Sum = 1 − cos 2β + cos 2α + cos 2β − 2cos²(α + β) + 2cos²(α + β) − 1 = cos 2α

Teacher's Note:
1. Put α = 0 for a quick check: 2 sin²β + cos 2β = 1 = cos 0.
2. 2 cos x cos y = cos (x + y) + cos (x − y) is the central tool here.

Answer: (b) real

Explanation:
1. G.P. gives sin²β = sin α cos α.
2. Discriminant = 4cot²β − 4 = 4(cos²β − sin²β)/sin²β = 4(1 − 2 sin²β)/sin²β
3. = 4(1 − sin 2α)/sin²β ≥ 0, so the roots are always real.

Teacher's Note:
1. Express the discriminant using the G.P. condition b² = ac.
2. 1 − sin 2α ≥ 0 because sin 2α ≤ 1.

Answer: (b) right angled

Explanation:
1. For A + B + C = π, sin²A + sin²B + sin²C = 2 + 2 cos A cos B cos C.
2. So 2 cos A cos B cos C = 0, i.e. one of the angles is 90°.

Teacher's Note:
1. This uses the conditional identity sin²A + sin²B + sin²C = 2 + 2 cos A cos B cos C.
2. A cosine of a triangle's angle is zero only for a right angle.

Answer: (d) 0

Explanation:
1. Write θ = (θ + α) − α and θ + 2α = (θ + α) + α.
2. sin (θ + α) cos α − cos (θ + α) sin α = 3 sin (θ + α) cos α + 3 cos (θ + α) sin α
3. −2 sin (θ + α) cos α = 4 cos (θ + α) sin α ⇒ tan (θ + α) = −2 tan α, so tan (θ + α) + 2 tan α = 0

Teacher's Note:
1. Rewrite both angles around the middle angle θ + α.
2. Dividing by cos (θ + α) cos α turns the result into tans.

Answer: (b) \( 4 - \sqrt{10} \)

Explanation:
1. cos²θ + 3 sin²θ = (1 + cos 2θ)/2 + 3(1 − cos 2θ)/2 = 2 − cos 2θ
2. Expression = 4 − cos 2θ − 3 sin 2θ
3. cos 2θ + 3 sin 2θ has maximum √10, so the least value is 4 − √10.

Teacher's Note:
1. Change squares and products into double angles to get a linear form in sin 2θ and cos 2θ.
2. a sin x + b cos x ranges between ±√(a² + b²).

Answer: (c) \( -\frac{3}{4} \)

Explanation:
1. sin³x = (3 sin x − sin 3x)/4.
2. sin α + sin (2π/3 + α) + sin (4π/3 + α) = 0, and sin 3x = sin 3α for each of the three angles.
3. Bracket = (3 × 0 − 3 sin 3α)/4 = −(3/4) sin 3α, so the value is −3/4

Teacher's Note:
1. The sines (or cosines) of three angles spaced 120° apart always add to 0.
2. Tripling an angle of 2π/3 + α gives 2π + 3α, which has the same sine as 3α.

Answer: (c) (−∞, 4]

Explanation:
1. k cos x − 3 sin x lies in [−√(k² + 9), √(k² + 9)].
2. Need (k + 1)² ≤ k² + 9 ⇒ 2k + 1 ≤ 9 ⇒ k ≤ 4.
3. So k ∈ (−∞, 4].

Teacher's Note:
1. a cos x + b sin x = c has a solution exactly when c² ≤ a² + b².
2. Squaring is safe here because both sides of the condition are non-negative.

Answer: (b) 6x² − x − 1 = 0

Explanation:
1. Put t = tan (θ/2): \( \frac{2t}{1 + t^2} + 7\cdot\frac{1 - t^2}{1 + t^2} = 5 \)
2. 2t + 7 − 7t² = 5 + 5t² ⇒ 12t² − 2t − 2 = 0
3. ⇒ 6t² − t − 1 = 0

Teacher's Note:
1. sin θ = 2t/(1 + t²) and cos θ = (1 − t²)/(1 + t²) with t = tan (θ/2).
2. Clear the common denominator 1 + t² before collecting terms.

Answer: (d) y = x − 3π/4

Explanation:
1. Expression = cos y sin x − sin y cos x + sin y sin x + cos x cos y = sin (x − y) + cos (x − y)
2. = √2 sin (x − y + π/4)
3. For y = x − 3π/4: x − y = 3π/4, so √2 sin π = 0. (Option (c) gives √2 sin (π/2) ≠ 0.)

Teacher's Note:
1. Replace each complementary angle first: cos (π/2 − x) = sin x, sin (π/2 − y) = cos y.
2. sin u + cos u = √2 sin (u + π/4) is zero when u + π/4 is a multiple of π.

Answer: (a) cos 12x = cos 14x; (b) sin 13x = 0; (c) sin x = 0

Explanation:
1. 2 cos 4x cos 8x = cos 12x + cos 4x and 2 cos 5x cos 9x = cos 14x + cos 4x.
2. Expression = (1/2)(cos 12x − cos 14x), which is 0 iff cos 12x = cos 14x.
3. cos 12x − cos 14x = 2 sin 13x sin x, so it is 0 if sin 13x = 0 or sin x = 0. cos x = 0 does not make it zero in general.

Teacher's Note:
1. Convert both products to sums; the common term cos 4x cancels.
2. Check option (d) with x = π/2: cos 2π cos 4π − cos (5π/2) cos (9π/2) = 1 ≠ 0.

Answer: (a) \( \theta \in \left(0, \frac{\pi}{2}\right) \); (b) \( \theta \in \left(\frac{\pi}{2}, \pi\right) \)

Explanation:
1. First term = sin²θ + sin θ cos θ + cos²θ = 1 + sin θ cos θ; last term = 2.
2. \( \sqrt{1 + \cot^2\theta} = |\text{cosec}\,\theta| \), so the middle term = cos θ |sin θ|.
3. LHS = −1 + sin θ cos θ − cos θ |sin θ|, which equals −1 when |sin θ| = sin θ, i.e. sin θ > 0.
4. sin θ > 0 for θ in (0, π/2) and (π/2, π) (θ = π/4, where sin θ = cos θ, is excluded as the expression is undefined there).

Teacher's Note:
1. √(1 + cot²θ) = |cosec θ|, not cosec θ; the sign decides the answer.
2. Sine is positive in the first and second quadrants.

Answer: (a) k₁ = 2k₂

Explanation:
1. \( \tan 3x - \tan x = \frac{\sin 3x\cos x - \cos 3x\sin x}{\cos 3x\cos x} = \frac{\sin 2x}{\cos 3x\cos x} = \frac{2\sin x}{\cos 3x} \)
2. So sin x/cos 3x = (1/2)(tan 3x − tan x).
3. k₂ = (1/2)[(tan 3θ − tan θ) + (tan 9θ − tan 3θ) + (tan 27θ − tan 9θ)] = (1/2)(tan 27θ − tan θ) = k₁/2

Teacher's Note:
1. Each term sin x/cos 3x is half a difference of tans, so the sum telescopes.
2. sin (3x − x) = sin 2x = 2 sin x cos x is the step that simplifies the numerator.

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

Explanation:
1. Put z = π/2 − y: sin x + sin z = a and cos x + cos z = b, so tan ((x + z)/2) = a/b.
2. (x + z)/2 = (x − y)/2 + π/4, so tan ((x − y)/2 + π/4) = a/b.
3. \( \tan\frac{x - y}{2} = \frac{\frac{a}{b} - 1}{1 + \frac{a}{b}} = \frac{a - b}{a + b} \)

Teacher's Note:
1. (sin x + sin z)/(cos x + cos z) = tan ((x + z)/2) after sum-to-product.
2. tan (u + π/4) = (1 + tan u)/(1 − tan u) can be inverted to find tan u.

Answer: (b) \( \pm\sqrt{\frac{3}{2}} \)

Explanation:
1. cot (θ − α) + cot (θ + α) = 6 cot θ ⇒ \( \frac{\sin 2\theta}{\sin(\theta - \alpha)\sin(\theta + \alpha)} = \frac{6\cos\theta}{\sin\theta} \)
2. sin (θ − α) sin (θ + α) = sin²θ − sin²α, so 2 sin²θ cos θ = 6 cos θ (sin²θ − sin²α).
3. cos θ ≠ 0 ⇒ sin²θ = 3 sin²θ − 3 sin²α ⇒ 2 sin²θ = 3 sin²α ⇒ sin θ cosec α = ±√(3/2)

Teacher's Note:
1. cot x + cot y = sin (x + y)/(sin x sin y).
2. sin (θ + α) sin (θ − α) = sin²θ − sin²α is a standard result worth remembering.

Answer: (a) sin²x − cos²x; (b) \( \sqrt{\frac{6}{5}}\left(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{3}}\cos x\right) \); (c) cos⁶x + sin⁶x; (d) cos²x + sin⁴x

Explanation:
1. (a) sin²x − cos²x = −cos 2x, maximum 1.
2. (b) maximum = √(6/5) × √(1/2 + 1/3) = √(6/5) × √(5/6) = 1.
3. (c) cos⁶x + sin⁶x = 1 − (3/4) sin²2x, maximum 1. (d) cos²x + sin⁴x = 1 − sin²x cos²x = 1 − (1/4) sin²2x, maximum 1.

Teacher's Note:
1. Reduce each function to a single trigonometric term whose range is known.
2. For a sin x + b cos x the maximum is √(a² + b²).

Answer: (a) \( \cos\frac{\theta - \phi}{2} = \pm\frac{1}{2}\sqrt{a^2 + b^2} \); (c) \( \tan\frac{\theta - \phi}{2} = \pm\sqrt{\frac{4 - a^2 - b^2}{a^2 + b^2}} \); (d) \( \cos(\theta - \phi) = \frac{a^2 + b^2 - 2}{2} \)

Explanation:
1. Squaring and adding: a² + b² = 2 + 2 cos (θ − φ), so cos (θ − φ) = (a² + b² − 2)/2. (d) is true.
2. a² + b² = 4 cos²((θ − φ)/2), so cos ((θ − φ)/2) = ±(1/2)√(a² + b²). (a) is true, (b) is false.
3. \( \tan^2\frac{\theta - \phi}{2} = \frac{1 - \cos(\theta - \phi)}{1 + \cos(\theta - \phi)} = \frac{4 - a^2 - b^2}{a^2 + b^2} \), so (c) is true.

Teacher's Note:
1. Squaring and adding sums of sines and cosines always produces cos (θ − φ).
2. Then the half-angle formulas give the other options.

Answer: (b) 8 : 5 : 9

Explanation:
1. In a triangle tan A + tan B + tan C = tan A tan B tan C, so 6 = 2 tan C ⇒ tan C = 3.
2. tan A + tan B = 3 and tan A tan B = 2 ⇒ {tan A, tan B} = {2, 1}; A > B ⇒ tan A = 2, tan B = 1.
3. sin² = tan²/(1 + tan²): 4/5, 1/2, 9/10, i.e. 8 : 5 : 9

Teacher's Note:
1. For A + B + C = π, tan A + tan B + tan C = tan A tan B tan C.
2. sin²x = tan²x/(1 + tan²x) avoids finding the angles.

Answer: (a) \( \tan\phi = \frac{1}{\sqrt{3}} \); (b) \( \tan\phi = -\frac{1}{\sqrt{3}} \); (c) \( \tan\theta = \sqrt{3} \); (d) \( \tan\theta = -\sqrt{3} \)

Explanation:
1. sin²θ = 3 sin²φ and sin θ cos φ = 3 cos θ sin φ ⇒ sin²θ cos²φ = 9 cos²θ sin²φ ⇒ cos²φ = 3 cos²θ.
2. sin²θ = 3(1 − cos²φ) = 3 − 9cos²θ ⇒ 1 − cos²θ = 3 − 9 cos²θ ⇒ cos²θ = 1/4, cos²φ = 3/4.
3. tan²θ = 3, tan²φ = 1/3 with tan θ = 3 tan φ: (tan θ, tan φ) = (√3, 1/√3) or (−√3, −1/√3). All four values can occur.

Teacher's Note:
1. Divide the two conditions to separate cos²φ and cos²θ.
2. Since only squares are fixed, both signs of tan are possible.

Answer: (a) \( \sqrt{2}\cos\left(\frac{\pi}{4} - x\right) \); (d) \( \sqrt{2}\sin\left(\frac{\pi}{4} + x\right) \)

Explanation:
1. \( \frac{\sin^3 x}{1 + \cos x} = \frac{\sin^3 x(1 - \cos x)}{\sin^2 x} = \sin x(1 - \cos x) \)
2. \( \frac{\cos^3 x}{1 - \sin x} = \frac{\cos^3 x(1 + \sin x)}{\cos^2 x} = \cos x(1 + \sin x) \)
3. Sum = sin x + cos x = √2 cos (π/4 − x) = √2 sin (π/4 + x)

Teacher's Note:
1. Multiply by the conjugate to replace 1 − cos²x by sin²x and 1 − sin²x by cos²x.
2. sin x + cos x can be written either as a sine or as a cosine of a shifted angle.

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

Explanation:
1. With u = θ/2: √(1 + sin θ) + √(1 − sin θ) = |sin u + cos u| + |sin u − cos u| = 2 max(|sin u|, |cos u|).
2. This equals 2 sin u exactly when sin u ≥ |cos u|, i.e. u ∈ [2nπ + π/4, 2nπ + 3π/4]. So A is true.
3. R is also true (sine is positive there), but positivity of sin u alone does not give the equation; the condition sin u ≥ |cos u| does. So R is not the correct explanation.

Teacher's Note:
1. Write 1 ± sin θ as (sin (θ/2) ± cos (θ/2))² to remove the square roots.
2. |p + q| + |p − q| = 2 max(|p|, |q|) is a useful fact for such expressions.

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

Explanation:
1. sin 18° = (√5 − 1)/4 and −sin 54° = −(√5 + 1)/4.
2. Sum = −1/2, product = −(5 − 1)/16 = −1/4, so both are roots of 4x² + 2x − 1 = 0. A is true.
3. R is true, and it explains A: 5x = 90° + 360°k gives sin 2x = cos 3x, which reduces to 4 sin²x + 2 sin x − 1 = 0 (after cancelling cos x); both 18° and −54° satisfy 5x ≡ 90° (mod 360°).

Teacher's Note:
1. 2x = 90° − 3x gives sin 2x = cos 3x, the standard route to the value of sin 18°.
2. Since −270° ≡ 90° (mod 360°), −54° satisfies the same condition as 18°.

Answer: (b) 3(m² − 1)²

Explanation:
1. P₁ = sin θ + cos θ = m ⇒ sin θ cos θ = (m² − 1)/2.
2. P₆ = 1 − 3 sin²θ cos²θ, so 4(1 − P₆) = 12 sin²θ cos²θ
3. = 12 × (m² − 1)²/4 = 3(m² − 1)²

Teacher's Note:
1. Square sin θ + cos θ to get the product sin θ cos θ.
2. sin⁶θ + cos⁶θ = 1 − 3 sin²θ cos²θ.

Answer: (c) 9

Explanation:
1. Let k = sin²θ cos²θ. P₆ = 1 − 3k and P₄ = 1 − 2k.
2. 2P₆ − 3P₄ + 10 = 2 − 6k − 3 + 6k + 10 = 9

Teacher's Note:
1. The k terms cancel, so the answer is a constant.
2. Check with θ = 0: 2(1) − 3(1) + 10 = 9.

Answer: (d) n − 4

Explanation:
1. Pₙ₋₂ − Pₙ = sinⁿ⁻²θ (1 − sin²θ) + cosⁿ⁻²θ (1 − cos²θ)
2. = sinⁿ⁻²θ cos²θ + cosⁿ⁻²θ sin²θ = sin²θ cos²θ (sinⁿ⁻⁴θ + cosⁿ⁻⁴θ)
3. = sin²θ cos²θ Pₙ₋₄, so λ = n − 4

Teacher's Note:
1. Factor out the lower power and use 1 − sin²θ = cos²θ.
2. Taking out sin²θ cos²θ lowers each power by 2 more.

Answer: (a) A→Q; B→R; C→S; D→P

Explanation:
1. (A) cos²x − sin²y = cos (x + y) cos (x − y) = cos 75° cos 30° = \( \frac{\sqrt{3} - 1}{2\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \frac{3 - \sqrt{3}}{4\sqrt{2}} \) → Q
2. (B) cos²108° + cos²144° = sin²18° + cos²36° = \( \frac{6 - 2\sqrt{5}}{16} + \frac{6 + 2\sqrt{5}}{16} = \frac{3}{4} \) → R
3. (C) sin 24° + sin 84° = 2 sin 54° cos 30° = \( 2 \cdot \frac{\sqrt{5} + 1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{15} + \sqrt{3}}{4} \) → S
4. (D) cos 130° = −cos 50°, so sin²50° + cos²50° = 1 → P

Teacher's Note:
1. cos²A − sin²B = cos (A + B) cos (A − B) is quick for (A).
2. Convert cos 6° into sin 84° so that sum-to-product applies in (C).

Answer: 2

Explanation:
1. 23° + 22° = 45°, so cot (23° + 22°) = 1.
2. \( \frac{\cot 23^\circ\cot 22^\circ - 1}{\cot 23^\circ + \cot 22^\circ} = 1 \) ⇒ cot 23° cot 22° − cot 23° − cot 22° = 1
3. (1 − cot 23°)(1 − cot 22°) = 1 − cot 23° − cot 22° + cot 23° cot 22° = 1 + 1 = 2

Teacher's Note:
1. Whenever A + B = 45°, (1 − cot A)(1 − cot B) = 2 and (1 + tan A)(1 + tan B) = 2.
2. Expand the product and compare with the cot addition formula.