JEE Mathematics Trigonometrical Functions and Identities MCQs Set 04

Answer: (c) \( \dfrac{2}{\sqrt5 + 1}\sqrt{10 - 2\sqrt5} \)

Explanation:
1. \( \tan63^\circ - \cot63^\circ = -2\cot126^\circ = 2\cot54^\circ \), and \( \cot54^\circ = \tan36^\circ \).
2. \( \tan36^\circ = \frac{\sin36^\circ}{\cos36^\circ} = \frac{\sqrt{10 - 2\sqrt5}/4}{(\sqrt5+1)/4} \), so the value is \( \frac{2}{\sqrt5+1}\sqrt{10 - 2\sqrt5} \approx 1.453 \).

Teacher's Note:
1. Use tan x - cot x = -2 cot 2x.
2. Use the exact values of sin 36° and cos 36°.

Answer: (c) 3

Explanation:
1. \( \cos^7x = 1 - \sin^4x \geq 0 \) and \( \cos^7x \leq |\cos x|^7 \); checking: \( \sin^4x = 1 - \cos^7x \) gives \( (1 - \cos^2x)^2 = 1 - \cos^7x \).
2. Solutions are \( \cos x = 1 \) (i.e. \( x = 0 \)) and \( \cos x = 0 \) (\( x = \pm\frac\pi2 \)). At \( x = \pm\pi \), \( \cos^7x = -1 \), which does not work. So there are 3 solutions.

Teacher's Note:
1. Test the special angles 0, ±π/2, ±π.
2. cos⁷x can be negative.

Answer: (a) 0

Explanation:
1. \( \tan70^\circ = \tan(50^\circ + 20^\circ) = \frac{\tan50^\circ + \tan20^\circ}{1 - \tan50^\circ\tan20^\circ} \).
2. Numerically \( 0.3640 + 2(1.1918) - 2.7475 = 0 \), and the identity \( \tan70^\circ = \tan20^\circ + 2\tan50^\circ \) holds exactly, so the value is 0.

Teacher's Note:
1. Use the identity tan 70° = tan 20° + 2 tan 50°.
2. Check numerically: 0.364 + 2.384 = 2.747.

Answer: (b) AP

Explanation:
1. AP gives \( \cos^4\theta\sec^2\alpha + \sin^4\theta\operatorname{cosec}^2\alpha = 1 \). By Cauchy-Schwarz the left side is at least \( \frac{(\cos^2\theta + \sin^2\theta)^2}{\cos^2\alpha + \sin^2\alpha} = 1 \), with equality only if \( \theta = \alpha \) (up to sign).
2. Then the new terms are \( \cos^2\alpha, \frac12, \sin^2\alpha \) whose first and last add to 1 = twice the middle: an AP.

Teacher's Note:
1. Cauchy-Schwarz forces equality.
2. Equality gives the relation θ = α.

Answer: (a) all four statements

Explanation:
1. Since \( \tan\frac\theta2(1 + \sec\theta) = \tan\theta \) (as \( 1 + \sec\theta = \frac{\cos\theta+1}{\cos\theta} \)), repeated use gives \( f_n(\theta) = \tan2^n\theta \).
2. \( f_2\left(\frac\pi{16}\right) = \tan\frac\pi4 = 1 \), \( f_3\left(\frac\pi{32}\right) = \tan\frac{8\pi}{32} = 1 \), \( f_4\left(\frac\pi{64}\right) = \tan\frac{16\pi}{64} = 1 \), \( f_5\left(\frac\pi{128}\right) = \tan\frac{32\pi}{128} = 1 \).

Teacher's Note:
1. Use tan(θ/2)(1 + sec θ) = tan θ repeatedly.
2. The product telescopes to tan 2ⁿθ.

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

Explanation:
1. The angles pair as \( \sin\frac{13\pi}{14} = \sin\frac{\pi}{14} \), \( \sin\frac{11\pi}{14} = \sin\frac{3\pi}{14} \), \( \sin\frac{9\pi}{14} = \sin\frac{5\pi}{14} \), and \( \sin\frac{7\pi}{14} = 1 \).
2. The product equals \( \left(\cos\frac\pi7\cos\frac{2\pi}7\cos\frac{3\pi}{7}\right)^2 \) with each \( \sin\frac{k\pi}{14} = \cos\frac{(7-k)\pi}{14} \), and \( \cos\frac\pi7\cos\frac{2\pi}7\cos\frac{3\pi}7 = \frac18 \). So the product is \( \frac{1}{64} \).

Teacher's Note:
1. Use sin(π - x) = sin x to pair terms.
2. cos(π/7) cos(2π/7) cos(3π/7) = 1/8.

Answer: (c) 0

Explanation:
1. \( 2^x + 2^{-x} \geq 2 \) by AM-GM, while \( \sin(e^x) \leq 1 \).
2. So the two sides never meet: no real solution.

Teacher's Note:
1. Compare the ranges of both sides.
2. AM-GM gives a minimum of 2.

Answer: (c) 2

Explanation:
1. \( \sin\theta + \frac{1}{\sin\theta} = 2 \) only when \( \sin\theta = 1 \).
2. Then \( \sin^8\theta + \operatorname{cosec}^8\theta = 1 + 1 = 2 \).

Teacher's Note:
1. t + 1/t = 2 only for t = 1.
2. Substitute and evaluate.

Answer: (d) \( [-1, +\infty) \)

Explanation:
1. With \( s = \sec\theta \), \( |s| \geq 1 \): \( \lambda = s^2 + s - 1 \).
2. For \( s \geq 1 \), \( \lambda \geq 1 \). For \( s \leq -1 \), \( \lambda \) takes all values from \( -1 \) (at \( s = -1 \)) upwards. So \( \lambda \in [-1, \infty) \).

Teacher's Note:
1. Use tan² = sec² - 1.
2. The vertex s = -1/2 is outside the allowed range.

Answer: (a) 6

Explanation:
1. \( \sin^3x = \frac{3\sin x - \sin3x}{4} \), so the product is \( \frac{3\sin x\sin3x - \sin^23x}{4} \).
2. \( \sin x\sin3x = \frac{\cos2x - \cos4x}{2} \) and \( \sin^23x = \frac{1 - \cos6x}{2} \). The highest multiple is \( 6x \), so \( n = 6 \).

Teacher's Note:
1. Reduce the power of sin x first.
2. The highest frequency decides n.

Answer: (b) A + B + C must be an integral multiple of \( \pi \)

Explanation:
1. \( \tan(A+B+C) = \frac{\sum\tan - \tan A\tan B\tan C}{1 - \sum\tan\tan} = 0 \).
2. So \( A + B + C = n\pi \) for an integer \( n \).

Teacher's Note:
1. The numerator of tan(A+B+C) vanishes.
2. tan x = 0 means x = nπ.

Answer: (b) \( 2x = y \)

Explanation:
1. \( x = \frac{\tan20^\circ + \tan50^\circ}{2} \) and \( y = \frac{\tan20^\circ + \tan70^\circ}{2} \).
2. Since \( \tan70^\circ = \tan(50^\circ + 20^\circ) \) satisfies \( \tan70^\circ = \tan20^\circ + 2\tan50^\circ \), \( y = \tan20^\circ + \tan50^\circ = 2x \).

Teacher's Note:
1. Use the identity tan 70° = tan 20° + 2 tan 50°.
2. Compare the two averages.

Answer: (a) \( \theta, \phi \)

Explanation:
1. \( x^2 + y^2 = r^2\sin^2\theta \), so \( x^2 + y^2 + z^2 = r^2\sin^2\theta + r^2\cos^2\theta = r^2 \).
2. It depends only on \( r \) and is independent of \( \theta \) and \( \phi \).

Teacher's Note:
1. These are spherical coordinates.
2. The sum of squares is the squared radius.

Answer: (d) \( \left(-\frac\pi2, \frac\pi2\right) \)

Explanation:
1. \( \sqrt{\frac{1 - \sin\alpha}{1 + \sin\alpha}} = \sqrt{\frac{(1-\sin\alpha)^2}{\cos^2\alpha}} = \left|\sec\alpha - \tan\alpha\right| \).
2. Equality needs \( \sec\alpha - \tan\alpha = \frac{1 - \sin\alpha}{\cos\alpha} \geq 0 \), i.e. \( \cos\alpha > 0 \). In \( [-\pi, \pi] \) that is \( \left(-\frac\pi2, \frac\pi2\right) \).

Teacher's Note:
1. Multiply numerator and denominator by 1 - sin α.
2. The root equals the absolute value.

Answer: (b) (i) and (ii)

Explanation:
1. \( 5\cos x + 12\sin x \in [-13, 13] \), so the denominator \( D \) has \( |D| \leq 13 \) and \( D \neq 0 \).
2. For \( D > 0 \): \( a = \frac1D \geq \frac{1}{13} \), the least positive value is \( \frac1{13} \). For \( D \lt 0 \): \( a \leq -\frac1{13} \), so the greatest negative value is \( -\frac{1}{13} \). (iii) and (iv) are false since \( a \) is unbounded.

Teacher's Note:
1. The reciprocal reverses the inequality.
2. Values between -1/13 and 1/13 are never reached.

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

Explanation:
1. \( a\cos\theta - b\sin\theta = R\cos(\theta + \delta) \) with \( R = \sqrt{a^2+b^2} \).
2. Its range is \( [-R, R] \).

Teacher's Note:
1. Write the expression as a single cosine.
2. Amplitude is √(a² + b²).

Answer: (d) none of these

Explanation:
1. We need \( \left|\frac{x^2-y^2}{x^2+y^2}\right| \geq 1 \), i.e. \( |x^2 - y^2| \geq x^2 + y^2 \).
2. This holds only when \( xy = 0 \) (not both zero): if \( y = 0 \), cosec \( \theta = 1 \); if \( x = 0 \), cosec \( \theta = -1 \). None of the options (a), (b), (c) describes this.

Teacher's Note:
1.
2. cosec θ
3. is at least 1.
4. The fraction has absolute value at most 1, with equality only when one variable is zero.

Answer: (c) 3

Explanation:
1. Using \( \tan\theta\tan(60^\circ+\theta)\tan(60^\circ-\theta) = \tan3\theta \) and the expansion in terms of \( t = \tan\theta \): the sum of the three tangents equals \( \frac{9t - 3t^3}{1 - 3t^2} = 3\tan3\theta \).
2. So \( k = 3 \). Numeric check at \( \theta = 10^\circ \): \( \tan10^\circ + \tan70^\circ + \tan(-50^\circ) = 1.732 = 3\tan30^\circ \).

Teacher's Note:
1. Write each tangent in terms of t = tan θ.
2. The sum simplifies to 3 tan 3θ.

Answer: (b) \( \cot\dfrac\pi{16} \)

Explanation:
1. Use \( \tan x = \cot x - 2\cot 2x \).
2. \( \tan\frac\pi{16} = \cot\frac\pi{16} - 2\cot\frac\pi8 \) and \( 2\tan\frac\pi8 = 2\cot\frac\pi8 - 4\cot\frac\pi4 = 2\cot\frac\pi8 - 4 \).
3. Adding: \( \tan\frac\pi{16} + 2\tan\frac\pi8 = \cot\frac\pi{16} - 4 \), so the required value is \( \cot\frac\pi{16} \).

Teacher's Note:
1. The identity cot x - tan x = 2 cot 2x telescopes.
2. Remember cot(π/4) = 1.

Answer: (c) 0

Explanation:
1. The sum of \( n \) terms of \( \sin\left[\frac\pi n + (k-1)\frac{2\pi}n\right] \) is \( \frac{\sin\left(n\cdot\frac\pi n\right)\sin\left(\frac\pi n + (n-1)\frac\pi n\right)}{\sin\frac\pi n} \).
2. This is \( \frac{\sin\pi\sin\pi}{\sin(\pi/n)} = 0 \).

Teacher's Note:
1. Use the formula for a sine series in AP.
2. sin π = 0.

Answer: (d) 4

Explanation:
1. \( \frac{\cos10^\circ - \sqrt3\sin10^\circ}{\sin10^\circ\cos10^\circ} = \frac{2\cos70^\circ}{\frac12\sin20^\circ} \).
2. \( \cos70^\circ = \sin20^\circ \), so the value is \( 4 \).

Teacher's Note:
1. Write √3/2 and 1/2 as sin 60° and cos 60°.
2. Use double angle for the denominator.

Answer: (d) 4

Explanation:
1. \( \frac{\sqrt3\cos20^\circ - \sin20^\circ}{\sin20^\circ\cos20^\circ} = \frac{2\sin(60^\circ - 20^\circ)}{\frac12\sin40^\circ} \).
2. \( = \frac{2\sin40^\circ}{\frac12\sin40^\circ} = 4 \).

Teacher's Note:
1. Combine into a single fraction and use √3 cos A - sin A = 2 sin(60° - A).
2. Use sin 2A = 2 sin A cos A in the denominator.

Answer: (b) \( 2\cos\dfrac{\theta}{2^n} \)

Explanation:
1. \( \sqrt{2(1+\cos\theta)} = 2\cos\frac\theta2 \) (the innermost radical, \( n = 1 \)).
2. Next, \( \sqrt{2 + 2\cos\frac\theta2} = 2\cos\frac\theta4 \). Each extra radical halves the angle, so for \( n \) radicals the value is \( 2\cos\frac{\theta}{2^n} \).

Teacher's Note:
1. Use 1 + cos A = 2cos²(A/2).
2. Each radical halves the angle.

Answer: (b) 12x^2 - 11x - 15 = 0

Explanation:
1. \( \sec A = -\frac54 \), so \( \cos A = -\frac45 \) and \( A \) is obtuse: \( \tan A = -\frac34 \), \( \operatorname{cosec}A = \frac53 \).
2. Sum \( = \frac{11}{12} \), product \( = -\frac54 \), so the equation is \( 12x^2 - 11x - 15 = 0 \).

Teacher's Note:
1. An angle of a convex quadrilateral is below 180°.
2. Decide the quadrant from the sign of cos A.

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

Explanation:
1. \( f(\theta) = \frac{1}{1 + \tan\theta} \).
2. \( \alpha + \beta = \frac{5\pi}{4} \) gives \( \tan(\alpha+\beta) = 1 \), so \( \tan\alpha + \tan\beta = 1 - \tan\alpha\tan\beta \), i.e. \( (1+\tan\alpha)(1+\tan\beta) = 2 \).
3. Hence \( f(\alpha)f(\beta) = \frac12 \).

Teacher's Note:
1. Rewrite f in terms of tan.
2. Use the addition formula for tan(α + β) = 1.

Answer: (a) 0

Explanation:
1. Opposite angles of a cyclic quadrilateral add to \( \pi \), so \( \cos\gamma = -\cos\alpha \) and \( \cos\delta = -\cos\beta \) (for the opposite pairs).
2. The sum is 0.

Teacher's Note:
1. Opposite angles are supplementary.
2. cos(π - x) = -cos x.