JEE Mathematics Trigonometrical Functions and Identities MCQs Set 05

Answer: (a) \( k_1=2k_2 \)

Explanation:
1. \( \tan3A-\tan A=\frac{\sin2A}{\cos A\cos3A}=\frac{2\sin A}{\cos3A} \), so \( \frac{\sin A}{\cos3A}=\frac12(\tan3A-\tan A) \).
2. Apply with A = θ, 3θ, 9θ and add: the terms telescope to \( k_2=\frac12(\tan27\theta-\tan\theta) \).
3. Hence k₁ = 2k₂.

Teacher's Note:
1. Prove the single-term identity tan 3A − tan A = 2 sin A / cos 3A first.
2. The three terms telescope.

Answer: (d) \( y-x=1 \)

Explanation:
1. \( \frac{99\pi}2=48\pi+\frac{3\pi}2 \), so θ lies in the fourth quadrant: sin θ < 0 and cos θ > 0.
2. Therefore \( x=-\sin^2\theta \) and \( y=\cos^2\theta \).
3. \( y-x=\cos^2\theta+\sin^2\theta=1 \).

Teacher's Note:
1. Find the quadrant of θ from the interval of values.
2. |sin θ| = −sin θ in the fourth quadrant.

Answer: (d) \( \cos1-\sin1\text{ and }\sqrt2 \)

Explanation:
1. Let \( u=\cos x\in[-1,1] \). Then \( A=\cos u+\sin u=\sqrt2\sin\left(u+\frac\pi4\right) \).
2. The argument \( u+\frac\pi4 \) ranges over \( \left[\frac\pi4-1,\frac\pi4+1\right]\approx[-0.215,1.785] \), which contains π/2, so the maximum is √2.
3. The minimum is at the lower end u = −1: \( \cos1-\sin1\approx-0.30 \).

Teacher's Note:
1. Reduce to a single sine and examine the range of its argument.
2. The range of the argument is not a full period, so the minimum is at an end point.

Answer: (b) \( 1+(p-r)^2 \)

Explanation:
1. Let \( f(x)=x^3-px^2-r=\prod(x-t_i) \), where \( t_i=\tan\alpha,\tan\beta,\tan\gamma \).
2. \( \prod(1+t_i^2)=\prod(t_i-i)(t_i+i)=f(i)\,f(-i) \).
3. \( f(i)=-i+p-r,\ f(-i)=i+p-r \), so the product is \( (p-r)^2+1 \).

Teacher's Note:
1. Write 1 + t² as (t − i)(t + i) and use the factored form of the cubic.
2. f(i) and f(−i) are complex conjugates.

Answer: (d) 36

Explanation:
1. \( \csc^2x+25\sec^2x=1+\cot^2x+25+25\tan^2x=26+\cot^2x+25\tan^2x \).
2. By AM-GM, \( \cot^2x+25\tan^2x\ge2\sqrt{25}=10 \).
3. The least value is 26 + 10 = 36.

Teacher's Note:
1. Expand using 1 + cot²x = csc²x and 1 + tan²x = sec²x.
2. Use AM-GM on cot²x and 25tan²x, whose product is constant.

Answer: (a) \( \frac34 \)

Explanation:
1. \( \csc^2\theta=1+\cot^2\theta=1+7=8 \) and \( \sec^2\theta=1+\frac17=\frac87 \).
2. \( \frac{8-\frac87}{8+\frac87}=\frac{48/7}{64/7}=\frac34 \).

Teacher's Note:
1. Use csc²θ = 1 + cot²θ and sec²θ = 1 + tan²θ.
2. Compute both values separately.

Answer: (b) \( \tan B-\tan C-\tan A \)

Explanation:
1. \( \tan B=\tan(A+C)=\frac{\tan A+\tan C}{1-\tan A\tan C} \).
2. Cross-multiply: \( \tan B-\tan A\tan B\tan C=\tan A+\tan C \).
3. So \( \tan A\tan B\tan C=\tan B-\tan A-\tan C \).

Teacher's Note:
1. Write B as A + C and use the addition formula.
2. Rearrange to isolate the triple product.

Answer: (d) \( \theta=\frac\pi3\text{ or }\frac{2\pi}3 \)

Explanation:
1. The infinite geometric series sums to \( \frac1{1-\sin\theta}=4+2\sqrt3 \).
2. \( 1-\sin\theta=\frac1{4+2\sqrt3}=\frac{4-2\sqrt3}{4}=1-\frac{\sqrt3}2 \), so \( \sin\theta=\frac{\sqrt3}2 \).
3. In (0, π), θ = π/3 or 2π/3.

Teacher's Note:
1. Use the sum S = 1/(1 − r) with r = sin θ.
2. Rationalise 1/(4 + 2√3).

Answer: (c) 1

Explanation:
1. From \( \sin x=1-\sin^2x=\cos^2x \).
2. \( \cos^8x+2\cos^6x+\cos^4x=\cos^4x(\cos^2x+1)^2 \).
3. With cos²x = sin x: \( \sin^2x(\sin x+1)^2=(\sin x+\sin^2x)^2=1 \).

Teacher's Note:
1. Factor the expression as a perfect square times cos⁴x.
2. Replace cos²x by sin x.

Answer: (a) \( \frac{\sqrt7-2}3 \)

Explanation:
1. Let \( t=\tan\frac\alpha2 \): \( \frac{2t+1-t^2}{1+t^2}=\frac{\sqrt7}2 \), i.e. \( (\sqrt7+2)t^2-4t+(\sqrt7-2)=0 \).
2. The discriminant is \( 16-4(7-4)=4 \), so \( t=\frac{4\pm2}{2(\sqrt7+2)} \): \( t=\frac{3}{\sqrt7+2}=\sqrt7-2 \) or \( t=\frac1{\sqrt7+2}=\frac{\sqrt7-2}3 \).
3. Since α < π/6, \( t<\tan15^\circ\approx0.268 \), which excludes \( \sqrt7-2\approx0.646 \). So \( t=\frac{\sqrt7-2}{3} \).

Teacher's Note:
1. Use the half-angle substitution for sin α and cos α.
2. Use the range of α to select one of the two roots.

Answer: (a) < 1

Explanation:
1. For u, v with sin u, sin v > 0: \( \sin(u+v)=\sin u\cos v+\cos u\sin v<\sin u+\sin v \) when cos u, cos v < 1.
2. Apply with u = α, v = β (cos < 1 strictly), giving \( \sin(\alpha+\beta)<\sin\alpha+\sin\beta \) and sin(α+β) > 0 for α + β < π.
3. Apply again with γ: \( \sin(\alpha+\beta+\gamma)<\sin(\alpha+\beta)+\sin\gamma<\sin\alpha+\sin\beta+\sin\gamma \). The ratio is less than 1.

Teacher's Note:
1. Sine is subadditive on [0, π]: sin(u + v) ≤ sin u + sin v.
2. Apply the property twice.

Answer: (c) \( \{c-b,\ b+c\} \)

Explanation:
1. The amplitude of \( P\sin x+Q\cos x \) is \( \sqrt{P^2+Q^2} \) with \( P=ab,\ Q=b\sqrt{1-a^2} \).
2. \( \sqrt{a^2b^2+b^2(1-a^2)}=b \).
3. So the expression ranges between c − b and c + b.

Teacher's Note:
1. Use the amplitude formula √(P² + Q²).
2. Add the constant c to the extremes.

Answer: (c) \( \frac45 \)

Explanation:
1. α is in the second quadrant (450° to 540° is 90° to 180° plus a full turn): \( \cos\alpha=-\frac{527}{625} \).
2. α/2 lies in (225°, 270°): \( \cos\frac\alpha2=-\sqrt{\frac{1+\cos\alpha}{2}}=-\sqrt{\frac{49}{625}}=-\frac7{25} \).
3. α/4 lies in (112.5°, 135°), where sine is positive: \( \sin\frac\alpha4=\sqrt{\frac{1-\cos(\alpha/2)}{2}}=\sqrt{\frac{16}{25}}=\frac45 \).

Teacher's Note:
1. Locate α, α/2, α/4 in their quadrants to fix the signs.
2. 625² − 336² = 527².

Answer: (b) −1

Explanation:
1. For fixed θ, the quadratic in x has its minimum at \( x=\frac{|\sin\theta|}{2} \): \( 4\cdot\frac{\sin^2\theta}{4}-4\cdot\frac{|\sin\theta|}2|\sin\theta|=-\sin^2\theta \).
2. Total \( -\sin^2\theta-\cos^2\theta=-1 \) for every θ.

Teacher's Note:
1. Complete the square in x: 4x² − 4x s = (2x − s)² − s².
2. The θ-dependence cancels.

Answer: (a) AP

Explanation:
1. AP gives \( \frac{\cos^4\theta}{\cos^2\alpha}+\frac{\sin^4\theta}{\sin^2\alpha}=1 \). Since \( \cos^2\alpha+\sin^2\alpha=1 \), this is equality in the Cauchy–Schwarz inequality, which forces \( \cos^2\theta=\cos^2\alpha \) and \( \sin^2\theta=\sin^2\alpha \).
2. Then \( \frac{\cos^8\theta}{\cos^6\alpha}+\frac{\sin^8\theta}{\sin^6\alpha}=\cos^2\theta+\sin^2\theta=1 \), so the new terms again sum to 1 around the middle term 1/2: they are in AP.

Teacher's Note:
1. Equality in (a²/x + b²/y)(x + y) ≥ (a + b)² forces a/x = b/y.
2. Substitute cos²θ = cos²α and sin²θ = sin²α in the second set.

Answer: (b) \( \sqrt2 \)

Explanation:
1. Since \( -1-\cos^2x=-2+\sin^2x \), put \( P=\sin^2x+2a^2 \): the expression is \( \sqrt P-\sqrt{P-2} \) with P ≥ 2 required for the second root to exist.
2. \( \sqrt P-\sqrt{P-2}=\frac2{\sqrt P+\sqrt{P-2}} \), which decreases as P increases.
3. Its maximum is at P = 2: \( \frac{2}{\sqrt2}=\sqrt2 \).

Teacher's Note:
1. Rewrite both radicands in terms of the same quantity P.
2. Rationalise to see that the difference decreases with P.