JEE Mathematics Trigonometric Functions MCQs Set 13

Answer: (c) \( -\frac{4+\sqrt7}{3} \)

Explanation:
1. Squaring: \( 1 + 2 \sin x \cos x = 1/4 \), so \( \sin x \cos x = -3/8 \) < 0. As 0 < \( x \) < \( \pi, \sin x \) > 0 and \( \cos x \) < 0.
2. \( (\sin x - \cos x)^{2} = 1 + 3/4 = 7/4 \), so \( \sin x - \cos x = \sqrt{7}/2 \).
3. Hence \( \sin x = (1 + \sqrt{7})/4 \) and \( \cos x = (1 - \sqrt{7})/4 \), so \( \tan x = (1 + \sqrt{7})/(1 - \sqrt{7}) = -(4 + \sqrt{7})/3 \).

Teacher's Note:
1. Use \( (\sin x + \cos x)^{2} \) and \( (\sin x - \cos x)^{2} \) together.
2. Decide the signs of \( \sin x \) and \( \cos x \) from the interval.

Answer: (a) x = y

Explanation:
1. \( \sec^{2}\theta \ge 1 \), while \( (x + y)^{2} \) ≥ 4xy gives 4xy/(x + y)² ≤ 1 when xy > 0.
2. So equality is needed: \( (x + y)^{2} \) = 4xy, i.e. \( (x - y)^{2} = 0 \), so \( x = y \) (non-zero).

Teacher's Note:
1. Compare the range of both sides.
2. AM–GM style inequality forces equality.

Answer: (b) 1

Explanation:
1. \( \tan^{2}\theta \cdot \cot^{2}\theta = 1 \) and \( \tan^{2}\theta \cdot \cos^{2}\theta = \sin^{2}\theta \).
2. So the expression = \( \sec^{2}\theta (1 - \sin^{2}\theta) = \sec^{2}\theta \cos^{2}\theta = 1 \).

Teacher's Note:
1. Multiply \( \tan^{2}\theta \) into the bracket first.
2. \( 1 - \sin^{2}\theta = \cos^{2}\theta \).

Answer: (a) \( 2\cot^n\frac{A-B}{2} \) if n is even

Explanation:
1. \( (\cos A + \cos B)/(\sin A - \sin B) = \cot((A - B)/2) \).
2. \( (\sin A + \sin B)/(\cos A - \cos B) = -\cot((A - B)/2) \).
3. The sum is \( \cot^n(1 + (-1)^n) \): equal to \( 2\cot^n((A - B)/2) \) for even \( n \) and 0 for odd \( n \).

Teacher's Note:
1. Use sum-to-product formulas for numerator and denominator.
2. \( (-1)^n \) decides between 2 and 0.

Answer: (a) π/6

Explanation:
1. Let \( t = 81^{\sin^{2}x} \); then \( 81^{\cos^{2}x} = 81/t \), so \( t + 81/t = 30 \), i.e. \( t^{2} - 30t + 81 = 0 \).
2. \( t = 3 \) or 27, so \( \sin^{2}x = 1/4 \) or 3/4.
3. \( x = \pi/6 \) satisfies \( \sin^{2}x = 1/4 \); none of the other options does.

Teacher's Note:
1. Use \( \sin^{2}x + \cos^{2}x = 1 \) to make the two exponents reciprocal.
2. Check each option in \( \sin^{2}x \).

Answer: (c) \( \frac12\left[1-\frac{\pi}{180}\right] \)

Explanation:
1. \( 1^{\circ} = \pi/180 \) radian. If the angles are \( x \) and \( y \) with \( x + y = 1 \) and \( x - y = \pi/180 \), then the smaller is \( y = (1 - \pi/180)/2 \).

Teacher's Note:
1. Convert the 1° difference to radians first.
2. Solve the two linear equations.

Answer: (a) cot α cot β cot γ

Explanation:
1. \( (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 \) for each angle.
2. Multiplying the two products gives \( 1 = \tan \alpha \tan \beta \tan \gamma \) × (required product).
3. So the required product = \( 1/(\tan \alpha \tan \beta \tan \gamma) = \cot \alpha \cot \beta \cot \gamma \).

Teacher's Note:
1. Use \( \sec^{2}\theta - \tan^{2}\theta = 1 \).
2. The two products are reciprocals of each other.

Answer: (a) \( (mn)^{-2/3} \)

Explanation:
1. \( m = \cos^{2}\theta/\sin \theta \) and \( n = \sin^{2}\theta/\cos \theta \), so mn = \( \sin \theta \cos \theta \).
2. \( m^{2/3} + n^{2/3} = \cos^{4/3}\theta/\sin^{2/3}\theta + \sin^{4/3}\theta/\cos^{2/3}\theta = (\cos^{2}\theta + \sin^{2}\theta)/(\sin \theta \cos \theta)^{2/3} \).
3. This equals \( 1/(mn)^{2/3} = (mn)^{-2/3} \).

Teacher's Note:
1. Simplify \( m \) and \( n \) to single fractions.
2. Combine over the common denominator \( (\sin \theta \cos \theta)^{2/3} \).

Answer: (b) \( f(x)=\sin^2x+\cos^2x;\ g(x)=1 \)

Explanation:
1. In \( (a), \sqrt{x^{2}} = |x| \) differs from \( x \) for \( x \) < 0. In \( (c), x/x \) is not defined at \( x = 0 \) but \( g(1) \) is defined everywhere.
2. In \( (b), \sin^{2}x + \cos^{2}x = 1 \) for every real \( x \), so the two functions have the same domain and values.

Teacher's Note:
1. Two functions are identical only if both domain and rule agree.

Answer: (b) π/4

Explanation:
1. At 4:30 the minute hand is at 180° from 12.
2. The hour hand is at \( 4 \times 30^{\circ} + 15^{\circ} = 135^{\circ} \).
3. The angle between them is \( 45^{\circ} = \pi/4 \).

Teacher's Note:
1. The hour hand moves 0.5° per minute.
2. Convert the final angle to radians.

Answer: (c) cos α cos β cos γ

Explanation:
1. Multiplying the two expressions: \( k^{2} = (1 - \sin^{2}\alpha)(1 - \sin^{2}\beta)(1 - \sin^{2}\gamma) = \cos^{2}\alpha \cos^{2}\beta \cos^{2}\gamma \).
2. For angles in \( [-\pi/2, \pi/2] \) each cosine is non-negative, so \( k = \cos \alpha \cos \beta \cos \gamma \).

Teacher's Note:
1. Multiply the two given products to get \( k^{2} \).
2. The sign is fixed by the interval of the angles.

Answer: (b) k = −sin²θ cos²θ

Explanation:
1. \( p_{n} \) − pₙ₋₂ = \( \cos^{n-2}\theta(\cos^{2}\theta - 1) + \sin^{n-2}\theta(\sin^{2}\theta - 1) \).
2. = \( -\sin^{2}\theta \cos^{n-2}\theta - \cos^{2}\theta \sin^{n-2}\theta = -\sin^{2}\theta \cos^{2}\theta (\cos^{n-4}\theta + \sin^{n-4}\theta) \).
3. So \( k = -\sin^{2}\theta \cos^{2}\theta \).

Teacher's Note:
1. Factor out the lower powers and use \( \cos^{2} - 1 = -\sin^{2} \).
2. Compare with pₙ₋₄.

Answer: (d) 0

Explanation:
1. Write \( \theta = (\theta + \alpha) - \alpha \) and \( \theta + 2\alpha = (\theta + \alpha) + \alpha \), and let \( \phi = \theta + \alpha \).
2. \( \sin(\phi - \alpha) = 3 \sin(\phi + \alpha) \) gives \( -2 \sin \phi \cos \alpha = 4 \cos \phi \sin \alpha \).
3. So \( \tan \phi = -2 \tan \alpha \), hence \( \tan(\theta + \alpha) + 2 \tan \alpha = 0 \).

Teacher's Note:
1. Rewrite both angles around θ + α.
2. Expand with the sum and difference formulas.

Answer: (d) 0

Explanation:
1. \( \cos 100^{\circ} + \cos 140^{\circ} = 2 \cos 120^{\circ} \cos 20^{\circ} = -\cos 20^{\circ} \).
2. So the sum is \( \cos 20^{\circ} - \cos 20^{\circ} = 0 \).

Teacher's Note:
1. Pair the two larger angles with the sum-to-product formula.

Answer: (c) 3 tan 3A

Explanation:
1. Using \( \tan(60^{\circ} + A) - \tan(60^{\circ} - A) = 2 \tan A \)... checking by a test value: \( A = 23^{\circ} \) gives 7.8153 and \( 3 \tan 69^{\circ} = 7.8153 \).
2. Algebraically, \( \tan A \tan(60^{\circ} - A) \tan(60^{\circ} + A) = \tan 3A \) and the combination equals \( 3 \tan 3A \).

Teacher's Note:
1. Test the formula with a convenient angle to confirm.
2. Remember the triple-angle identity \( \tan A \tan(60^{\circ} - A) \tan(60^{\circ} + A) = \tan 3A \).

Answer: (a) 1

Explanation:
1. The product has 17 factors (5°, 10°, ..., 85°).
2. \( \cot x \cot(90^{\circ} - x) = 1 \), so the factors pair off (5° with 85°, 10° with 80°, ...) to give 1 each.
3. The middle factor is \( \cot 45^{\circ} = 1 \), so the product is 1.

Teacher's Note:
1. Pair complementary angles.
2. The unpaired middle factor is \( \cot 45^{\circ} \).

Answer: (c) 1 and 3 only

Explanation:
1. \( \sin A + \sin 3A = 2 \sin 2A \cos A \), so the sum is \( 2 \sin 2A (\cos A + 1) = 2 \sin 2A \cdot 2\cos^{2}(A/2) = 4 \sin 2A \cos^{2}(A/2) \). This is statement 1.
2. Statement 3 is the same since \( \sin 2A = 2 \sin A \cos A \).
3. Statement 2 equals \( 2 \sin 2A (1 + \sin A) \), which is not the same.

Teacher's Note:
1. Group \( \sin A \) and \( \sin 3A \) first.
2. \( 1 + \cos A = 2\cos^{2}(A/2) \).

Answer: (d) 1/8

Explanation:
1. \( \cos(5\pi/8) = -\cos(3\pi/8) \) and \( \cos(7\pi/8) = -\cos(\pi/8) \).
2. The product = \( (1 - \cos^{2}(\pi/8))(1 - \cos^{2}(3\pi/8)) = \sin^{2}(\pi/8) \sin^{2}(3\pi/8) \).
3. Since \( \sin(3\pi/8) = \cos(\pi/8) \), this is \( (\sin(\pi/8)\cos(\pi/8))^{2} = (\sin(\pi/4)/2)^{2} = 1/8 \).

Teacher's Note:
1. Replace the obtuse angles using \( \cos(\pi - x) = -\cos x \).
2. Use the double-angle formula at the end.

Answer: (b) −cosec 2°

Explanation:
1. \( \cot 224^{\circ} = \cot 44^{\circ}, \cot 134^{\circ} = -\tan 44^{\circ}, \cot 226^{\circ} = \tan 44^{\circ}, \cot 316^{\circ} = -\cot 44^{\circ} \).
2. The ratio = \( (\cot 44^{\circ} + \tan 44^{\circ})/(\tan 44^{\circ} - \cot 44^{\circ}) = 1/(\sin^{2}44^{\circ} - \cos^{2}44^{\circ}) × ... = 1/(-\cos 88^{\circ}) = -\sec 88^{\circ} = -\csc 2^{\circ} \).

Teacher's Note:
1. Reduce every angle to 44° using the quadrant rules.
2. \( \cot x + \tan x = 1/(\sin x \cos x) \) and \( \tan x - \cot x = (\sin^{2}x - \cos^{2}x)/(\sin x \cos x) \).

Answer: (d) 1/2, −1/2

Explanation:
1. \( B = 90^{\circ} - A \), so \( \cos A \cos B = \cos A \sin A = (1/2) \sin 2A \).
2. \( \sin 2A \) varies between −1 and 1, so the greatest and least values are 1/2 and −1/2.

Teacher's Note:
1. Write the product as a single sine.
2. The range of \( \sin 2A \) is [−1, 1].

Answer: (b) 0

Explanation:
1. \( \sin^{6}\theta + \cos^{6}\theta = 1 - 3 \sin^{2}\theta \cos^{2}\theta \) and \( \sin^{4}\theta + \cos^{4}\theta = 1 - 2 \sin^{2}\theta \cos^{2}\theta \).
2. The expression = \( 2 - 6s - 3 + 6s + 1 = 0 \) where \( s = \sin^{2}\theta \cos^{2}\theta \).

Teacher's Note:
1. Use the identities for \( a^{3} + b^{3} \) and \( a^{2} + b^{2} \) with \( a = \sin^{2}\theta, b = \cos^{2}\theta \).

Answer: (a) x² + y² − z²

Explanation:
1. \( (x \cos \theta + y \sin \theta)^{2} + (x \sin \theta - y \cos \theta)^{2} = x^{2} + y^{2} \).
2. So \( (x \sin \theta - y \cos \theta)^{2} = x^{2} + y^{2} - z^{2} \).

Teacher's Note:
1. Add and subtract the two squares to eliminate the cross terms.

Answer: (d) None of these

Explanation:
1. \( \sin x - \sin 3x = -2 \cos 2x \sin x \) and \( \sin^{2}x - \cos^{2}x = -\cos 2x \).
2. The ratio = \( 2 \sin x \), which is none of the three given expressions.

Teacher's Note:
1. Use the difference-to-product formula.
2. Check whether the simplified result matches an option.

Answer: (b) 2(1 − sin A)(1 + cos A)

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

Teacher's Note:
1. Expand carefully and group by \( (1 - \sin A) \).

Answer: (b) 2 cos x

Explanation:
1. Numerator = \( 32 \cos^{6}x \) and denominator = \( 16 \cos^{5}x \), because \( \cos^{6}x \) and \( \cos^{5}x \) expand into these multiple-angle cosines.
2. So the ratio = \( 2 \cos x \).

Teacher's Note:
1. Recognise the binomial-type coefficients 1, 6, 15, 10 and 1, 5, 10.
2. Check with \( x = 0 \): \( (1 + 6 + 15 + 10)/(1 + 5 + 10) = 2 \).