JEE Mathematics Trigonometrical Functions and Identities MCQs Set 03

Answer: (c) \( a^2 + b^2 = c^2 + d^2 \)

Explanation:
1. Rewrite as \( a\sec\alpha - c\tan\alpha = d \) and \( b\sec\alpha + d\tan\alpha = c \). Square and add after moving terms: \( (a\sec\alpha)^2 + (b\sec\alpha)^2 = (d + c\tan\alpha)^2 + (c - d\tan\alpha)^2 \).
2. The right side is \( (c^2 + d^2)(1 + \tan^2\alpha) = (c^2+d^2)\sec^2\alpha \). Cancelling \( \sec^2\alpha \): \( a^2 + b^2 = c^2 + d^2 \).

Teacher's Note:
1. Isolate a sec α and b sec α, then square and add.
2. 1 + tan² α = sec² α.

Answer: (a) 0

Explanation:
1. With \( s = \sin18^\circ \) and \( c = \cos18^\circ \): the expression is \( \frac{2s + 3 - 4c^2}{c} = \frac{4s^2 + 2s - 1}{c} \).
2. \( s = \frac{\sqrt5 - 1}{4} \) satisfies \( 4s^2 + 2s - 1 = 0 \), so the value is 0.

Teacher's Note:
1. Put everything over cos 18°.
2. sin 18° is a root of 4s² + 2s - 1 = 0.

Answer: (c) \( \dfrac12\left(\dfrac xy + \dfrac yx\right) \)

Explanation:
1. Then \( e^{i\alpha} = x \) (or \( \frac1x \)) and \( e^{i\beta} = y \) (or \( \frac1y \)), taking \( \alpha, \beta \) as complex angles.
2. \( \cos(\alpha - \beta) = \frac12\left(e^{i(\alpha-\beta)} + e^{-i(\alpha-\beta)}\right) = \frac12\left(\frac xy + \frac yx\right) \).

Teacher's Note:
1. cos θ = (e^iθ + e^-iθ)/2 matches the given form.
2. Apply it to the difference.

Answer: (c) HP

Explanation:
1. Expand: \( 2\sin\alpha\cos\beta\sin\gamma = \sin\beta\sin\alpha\cos\gamma + \sin\beta\cos\alpha\sin\gamma \).
2. Divide by \( \sin\alpha\sin\beta\sin\gamma \): \( 2\cot\beta = \cot\gamma + \cot\alpha \). The cotangents are in AP, so the tangents are in HP.

Teacher's Note:
1. Divide by the product of sines.
2. An AP of cotangents means an HP of tangents.

Answer: (a) 0

Explanation:
1. \( (3\sin\theta + 4\cos\theta)^2 + (4\sin\theta - 3\cos\theta)^2 = 25 \).
2. The first square is 25, so \( 4\sin\theta - 3\cos\theta = 0 \).

Teacher's Note:
1. The sum of the squares is always 25.
2. Use the identity (a sin + b cos)² + (b sin - a cos)² = a² + b².

Answer: (c) 1

Explanation:
1. With \( c = \cos x \): \( 2c^2 + 2c - 2 = 0 \), so \( c^2 + c - 1 = 0 \) and \( c^2 = 1 - c \).
2. \( \sin^2x(2 - c^2) = (1 - c^2)(2 - c^2) = c(1 + c) = c + c^2 = 1 \).

Teacher's Note:
1. Get the quadratic in cos x.
2. Substitute c² = 1 - c.

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

Explanation:
1. \( \tan A\tan B = \frac{\cos(A-B) - \cos(A+B)}{\cos(A-B) + \cos(A+B)} = \frac{c - \frac12}{c + \frac12} \) with \( c = \cos(A-B) \in \left[\frac12, 1\right] \).
2. This increases with \( c \), so the maximum (at \( A = B \)) is \( \frac{1/2}{3/2} = \frac13 \); the minimum is 0 (at \( A = 0 \)).

Teacher's Note:
1. Use the product-to-sum formula for tan A tan B.
2. Find the range of cos(A - B).

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

Explanation:
1. Sum \( = \frac{13}{4} \) and product \( = \frac{15}{8} \), so \( \tan\frac{\alpha+\beta}{2} = \frac{13/4}{1 - 15/8} = -\frac{26}{7} \).
2. \( \cos(\alpha+\beta) = \frac{1 - t^2}{1 + t^2} = \frac{49 - 676}{49 + 676} = -\frac{627}{725} \).

Teacher's Note:
1. Find tan of the half-sum first.
2. Then use the half-angle form of cos.

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

Explanation:
1. \( (\cos20^\circ + \sin20^\circ)^2 = 2 - (\cos20^\circ - \sin20^\circ)^2 = 2 - p^2 \), so \( \cos20^\circ + \sin20^\circ = \sqrt{2 - p^2} \) (positive).
2. \( \cos40^\circ = \cos^220^\circ - \sin^220^\circ = p\sqrt{2 - p^2} \).

Teacher's Note:
1. Use cos 2A = (cos A - sin A)(cos A + sin A).
2. (a + b)² + (a - b)² = 2(a² + b²).

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

Explanation:
1. Multiply by \( 2\sin\frac\pi{11} \): each \( 2\sin\frac\pi{11}\cos\frac{(2k-1)\pi}{11} = \sin\frac{2k\pi}{11} - \sin\frac{(2k-2)\pi}{11} \), which telescopes to \( \sin\frac{10\pi}{11} = \sin\frac\pi{11} \).
2. So the sum is \( \frac12 \).

Teacher's Note:
1. Multiply by 2 sin(π/11) to telescope.
2. sin(10π/11) = sin(π/11).

Answer: (d) \( -\frac12 \)

Explanation:
1. \( \sin\theta + \sin3\theta = 2\sin2\theta\cos\theta \), so the equation is \( \sin2\theta(2\cos\theta + 1) = 0 \).
2. Since \( \sin2\theta \neq 0 \), \( \cos\theta = -\frac12 \).

Teacher's Note:
1. Combine sin θ + sin 3θ.
2. Factor out sin 2θ.

Answer: (b) \( \frac{-9}{8} \)

Explanation:
1. With \( c = \cos\theta \): \( \cos2\theta + \cos\theta = 2c^2 + c - 1 \).
2. The vertex is at \( c = -\frac14 \), within \( [-1, 1] \), giving \( 2\cdot\frac1{16} - \frac14 - 1 = -\frac98 \).

Teacher's Note:
1. Convert to a quadratic in cos θ.
2. Check the vertex lies in [-1, 1].

Answer: (a) \( 4\sin A\sin B\sin C \)

Explanation:
1. \( \sin2A + \sin2B = 2\sin(A+B)\cos(A-B) = 2\sin C\cos(A-B) \) and \( \sin2C = 2\sin C\cos C = -2\sin C\cos(A+B) \).
2. Sum \( = 2\sin C[\cos(A-B) - \cos(A+B)] = 4\sin A\sin B\sin C \).

Teacher's Note:
1. Use A + B = π - C.
2. Apply product-to-sum on cosines.

Answer: (c) a negative rational number

Explanation:
1. Multiply by \( 2\sin\frac\pi7 \): the sum telescopes to \( \sin\frac{7\pi}7 - \sin\frac\pi7 = -\sin\frac\pi7 \).
2. So the sum is \( -\frac12 \), a negative rational number.

Teacher's Note:
1. Multiply by 2 sin(π/7).
2. sin π = 0.

Answer: (b) (ii) and (iii)

Explanation:
1. \( \cos(x - y) = -1 \) means \( x - y = (2k+1)\pi \), so \( x = y + (2k+1)\pi \).
2. Then \( \cos x = -\cos y \) and \( \sin x = -\sin y \), which are statements (ii) and (iii).

Teacher's Note:
1. cos = -1 means the angle is an odd multiple of π.
2. Shifting by π negates both sine and cosine.

Answer: (c) \( \left[-2, \frac98\right] \)

Explanation:
1. \( k = -(2c^2 + c - 1) \) with \( c = \cos\theta \in [-1, 1] \).
2. \( 2c^2 + c - 1 \) ranges over \( \left[-\frac98, 2\right] \), so \( k \in \left[-2, \frac98\right] \).

Teacher's Note:
1. Convert to a quadratic in cos θ.
2. Find its range on [-1, 1].

Answer: (b) \( \cos2\alpha \)

Explanation:
1. \( \sin2\alpha = \frac{2\sqrt a}{1+a} \) and \( \tan2\alpha = \frac{2\sqrt a}{1-a} \) are irrational.
2. \( \cos2\alpha = \frac{1 - \tan^2\alpha}{1 + \tan^2\alpha} = \frac{1-a}{1+a} \) is rational.

Teacher's Note:
1. Express each in terms of tan α.
2. Only the even function cos 2α avoids the odd power √a.

Answer: (b) \( -\frac12 \)

Explanation:
1. Group as \( (\sin78^\circ + \sin6^\circ) - (\sin66^\circ + \sin42^\circ) = 2\sin42^\circ\cos36^\circ - 2\sin54^\circ\cos12^\circ \).
2. \( \sin54^\circ = \cos36^\circ \) and \( \cos12^\circ = \sin78^\circ \), so the value is \( 2\cos36^\circ(\sin42^\circ - \sin78^\circ) = -4\cos36^\circ\cos60^\circ\sin18^\circ = -2\cos36^\circ\sin18^\circ \).
3. \( \cos36^\circ\sin18^\circ = \frac14 \), so the value is \( -\frac12 \).

Teacher's Note:
1. Use sum-to-product on pairs.
2. The values of sin 18° and cos 36° are known exactly.

Answer: (b) p^2 = q(q + 2)

Explanation:
1. \( \sec\alpha + \operatorname{cosec}\alpha = p \) and \( \sec\alpha\operatorname{cosec}\alpha = q \), so \( \sin\alpha\cos\alpha = \frac1q \).
2. \( p^2 = \frac{(\sin\alpha + \cos\alpha)^2}{\sin^2\alpha\cos^2\alpha} = q^2\left(1 + \frac2q\right) = q(q + 2) \).

Teacher's Note:
1. Express p² in sin α and cos α.
2. Use 2 sin α cos α = 2/q.

Answer: (a) (i) and (iv)

Explanation:
1. This is the Ptolemy-type identity \( \sin(\alpha-\beta)\sin(\gamma-\delta) + \sin(\beta-\gamma)\sin(\alpha-\delta) + \sin(\gamma-\alpha)\sin(\beta-\delta) = 0 \), so (i) holds.
2. If \( x + y + z = 0 \) then \( x^3 + y^3 + z^3 = 3xyz \), so (iv) holds as well. (ii) and (iii) fail in general.

Teacher's Note:
1. If a + b + c = 0 then a³ + b³ + c³ = 3abc.
2. Check the identity with simple angles.

Answer: (a) \( \left(-\frac54, 1\right] \)

Explanation:
1. The union is the set where \( \sin\theta > -\frac12 \). Also \( \sin\theta = 1 - \sqrt{1 - \lambda} \) requires \( \lambda \leq 1 \).
2. \( 1 - \sqrt{1-\lambda} > -\frac12 \) gives \( \sqrt{1 - \lambda} \lt \frac32 \), so \( \lambda > -\frac54 \). Hence \( \lambda \in \left(-\frac54, 1\right] \).

Teacher's Note:
1. Solve the quadratic for sin θ.
2. The interval means sin θ > -1/2.

Answer: (c) \( \frac n2 - 1 \)

Explanation:
1. \( \cos^2\theta = \frac{1 + \cos2\theta}{2} \), so the sum is \( \frac{n-1}{2} + \frac12\sum_{r=1}^{n-1}\cos\frac{2r\pi}{n} \).
2. \( \sum_{r=1}^{n-1}\cos\frac{2r\pi}{n} = -1 \) (all n-th roots of unity add to 0). The sum is \( \frac{n-1}{2} - \frac12 = \frac n2 - 1 \).

Teacher's Note:
1. Use the half-angle identity.
2. Roots of unity sum to zero.

Answer: (b) \( \tan3\theta \)

Explanation:
1. This is the standard identity: \( \tan\theta\tan(60^\circ - \theta)\tan(60^\circ + \theta) = \tan3\theta \).
2. Check with \( \theta = 15^\circ \): \( \tan15^\circ\tan45^\circ\tan75^\circ = 1 = \tan45^\circ \).

Teacher's Note:
1. Memorise this identity.
2. Verify with θ = 15°.

Answer: (d) \( \frac{-1}{3} \)

Explanation:
1. \( \cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} = 2\cdot\frac{\sqrt3}{2}\cos\frac{A-B}{2} = 1 \), so \( \cos\frac{A-B}{2} = \frac{1}{\sqrt3} \).
2. \( \cos(A - B) = 2\cos^2\frac{A-B}{2} - 1 = \frac23 - 1 = -\frac13 \).

Teacher's Note:
1. Use sum-to-product with A + B known.
2. Then apply the double-angle formula.

Answer: (b) 39x^2 - 16x - 48 = 0

Explanation:
1. \( \tan A = \frac5{12} \) with \( A \) acute gives \( \cos A = \frac{12}{13} \), so \( \cos C = \cos(\pi - A) = -\frac{12}{13} \).
2. \( \cos B = -\frac35 \) makes \( B \) obtuse with \( \tan B = -\frac43 \), so \( \tan D = \tan(\pi - B) = \frac43 \).
3. Sum \( = \frac{16}{39} \), product \( = -\frac{48}{39} \): \( 39x^2 - 16x - 48 = 0 \).

Teacher's Note:
1. In a cyclic quadrilateral C = π - A and D = π - B.
2. Pick the correct quadrants first.

Answer: (b) 0

Explanation:
1. Let \( t = e^{\sin x} \): \( t - \frac1t = 4 \), so \( t = 2 + \sqrt5 \).
2. \( \sin x = \ln(2 + \sqrt5) \approx 1.44 > 1 \), which is impossible. So there are no real values.

Teacher's Note:
1. Substitute t = e^(sin x).
2. Check the value against the range of sine.