JEE Mathematics Trigonometrical Functions and Identities MCQs Set 06

Answer: (a) \( \frac{\sin A}{(1-n)\cos A} \)

Explanation:
1. Numerator \( \tan A+\tan B=\frac{\sin A}{\cos A}+\frac{n\sin A\cos A}{1-n\cos^2A}=\frac{\sin A}{\cos A(1-n\cos^2A)} \).
2. Denominator \( 1-\tan A\tan B=1-\frac{n\sin^2A}{1-n\cos^2A}=\frac{1-n}{1-n\cos^2A} \).
3. \( \tan(A+B)=\frac{\sin A}{(1-n)\cos A} \).

Teacher's Note:
1. Use tan(A + B) = (tan A + tan B)/(1 − tan A tan B) and combine the fractions.
2. Note 1 − n cos²A − n sin²A = 1 − n.

Answer: (b) \( \sqrt2 \)

Explanation:
1. Let \( \omega=e^{2\pi i/7} \). Then \( B+iA=\omega+\omega^2+\omega^4 \) (the sum of three of the primitive 7th roots of unity).
2. Call it \( S \) and the conjugate sum \( \bar S=\omega^6+\omega^5+\omega^3 \). Then \( S+\bar S=-1 \) and \( S\bar S=2 \) (expand and use \( 1+\omega+\dots+\omega^6=0 \)).
3. \( A^2+B^2=|S|^2=S\bar S=2 \), so \( \sqrt{A^2+B^2}=\sqrt2 \). (Numerically \( B=-\frac12,\ A=\frac{\sqrt7}2 \).)

Teacher's Note:
1. Treat the two sums as the real and imaginary parts of a sum of 7th roots of unity.
2. Use the fact that the sum of all 7th roots of unity is zero.

Answer: (b) \( \left\{\frac\pi{12},\frac{7\pi}{12},\frac{17\pi}{12},\frac{23\pi}{12}\right\} \)

Explanation:
1. \( \cos\theta=\frac{2\sqrt2\pm\sqrt{8+16}}{8}=\frac{\sqrt2\pm\sqrt6}{4} \).
2. \( \cos\theta=\frac{\sqrt6+\sqrt2}{4}=\cos15^\circ \) gives \( \theta=\frac\pi{12},\frac{23\pi}{12} \).
3. \( \cos\theta=\frac{\sqrt2-\sqrt6}{4}=-\cos75^\circ=\cos105^\circ \) gives \( \theta=\frac{7\pi}{12},\frac{17\pi}{12} \).

Teacher's Note:
1. Recognise (√6 + √2)/4 = cos 15° and (√6 − √2)/4 = cos 75°.
2. Each value of cos θ gives two angles in [0, 2π].

Answer: (a) only III is false

Explanation:
1. Put \( a=\cos\varphi,\ b=\sin\varphi \). The equation becomes \( \sin(x+\varphi)=1 \), so \( x=\frac\pi2-\varphi+2n\pi \).
2. \( \sin x=\cos\varphi=a \) (I true) and \( \tan x=\cot\varphi=\frac ab \) (II true).
3. \( \tan x=\frac ab\ne b \) in general (III false).

Teacher's Note:
1. Because a² + b² = 1 you can treat (a, b) as (cos φ, sin φ).
2. Test each statement using x = π/2 − φ.

Answer: (a) \( \frac73 \)

Explanation:
1. \( \frac{\cos3x}{\cos x}=4\cos^2x-3=\frac13\Rightarrow\cos^2x=\frac56 \), so \( \sin^2x=\frac16 \).
2. \( \frac{\sin3x}{\sin x}=3-4\sin^2x=3-\frac46=\frac73 \).

Teacher's Note:
1. Use cos 3x = cos x (4cos²x − 3) and sin 3x = sin x (3 − 4 sin²x).
2. As a check, sin 3x / sin x − cos 3x / cos x = 2 for every x, and 7/3 − 1/3 = 2.

Answer: (d) \( -\tan\frac x2 \)

Explanation:
1. \( \sqrt{1\mp\sin x}=\left|\sin\frac x2\mp\cos\frac x2\right| \). Here \( \frac{5\pi}4\lt \frac x2\lt \frac{3\pi}2 \), so both sine and cosine are negative and \( |\sin\frac x2|\gt |\cos\frac x2| \).
2. \( \sqrt{1-\sin x}=\cos\frac x2-\sin\frac x2 \) and \( \sqrt{1+\sin x}=-\left(\sin\frac x2+\cos\frac x2\right) \).
3. Numerator \( =-2\sin\frac x2 \), denominator \( =2\cos\frac x2 \), ratio \( =-\tan\frac x2 \).

Teacher's Note:
1. Use 1 ± sin x = (sin(x/2) ± cos(x/2))² and decide the sign of each bracket from the quadrant of x/2.
2. x/2 lies in the third quadrant, closer to 3π/2.

Answer: (d) \( \frac{3-9\tan^2x}{3\tan x-\tan^3x} \)

Explanation:
1. Using \( \cot x+\cot(60^\circ+x)+\cot(120^\circ+x)=3\cot3x \) (check at \( x=0.3 \) rad: both sides equal 2.3807).
2. \( 3\cot3x=\frac{3(1-3\tan^2x)}{3\tan x-\tan^3x}=\frac{3-9\tan^2x}{3\tan x-\tan^3x} \).

Teacher's Note:
1. Recall tan x tan(60° − x) tan(60° + x) = tan 3x and the related sum of cotangents.
2. Convert cot 3x into tan x using tan 3x = (3t − t³)/(1 − 3t²).

Answer: (a) −1

Explanation:
1. Write \( \cos2x=2\cos^2x-1 \). Then \( y=\left(\frac{x}{2\sqrt2}+\sqrt2\cos x\right)^2-1 \).
2. A square is non-negative, so \( y\ge-1 \).
3. The value \( -1 \) is attained when \( x=-4\cos x \) (this has a real solution), so the best lower bound is −1.

Teacher's Note:
1. Complete the square using cos 2x = 2cos²x − 1.
2. Check that the bound is actually reached.

Answer: (a) \( \frac{3}{\sin\theta\cos^2\theta} \)

Explanation:
1. Let T be the top-right corner, B the bottom-right corner (folded) and Q its image on the left side. The crease is TR with R on the bottom side, \( \angle BTR=\theta \) and the right angle at B, so \( TB=l\cos\theta \).
2. Folding preserves lengths and angles: \( TQ=TB=l\cos\theta \) and \( \angle RTQ=\theta \), so \( \angle BTQ=2\theta \).
3. The horizontal distance from T to the left side is 6: \( TQ\sin2\theta=6\Rightarrow l\cos\theta\cdot2\sin\theta\cos\theta=6\Rightarrow l=\frac{3}{\sin\theta\cos^2\theta} \).

Teacher's Note:
1. Draw the folded triangle: the crease bisects the angle between the original and folded positions of the side.
2. The horizontal projection of the folded side equals the width 6.

Answer: (a) \( \left(0,\frac\pi4\right) \)

Explanation:
1. On \( \left(0,\frac\pi4\right) \): \( \sin x\lt \cos x \) and \( \tan x\lt \cot x \) hold, and \( \cos x\lt \tan x \) holds once \( x \) exceeds about 0.666 rad (that is, \( \sin x\gt \frac{\sqrt5-1}{2} \)), which is less than \( \frac\pi4\approx0.785 \). So the chain is possible there.
2. In \( \left(\frac{3\pi}4,\pi\right) \) sin is positive and cos negative, so \( \sin x\lt \cos x \) fails. In \( \left(\frac{5\pi}4,\frac{3\pi}2\right) \) \( \tan x\gt 1\gt \cot x \) so \( \tan x\lt \cot x \) fails. In \( \left(\frac{7\pi}4,2\pi\right) \) tan is negative and below cos, so \( \cos x\lt \tan x \) fails.

Teacher's Note:
1. Test each interval against the individual inequalities in the chain.
2. Find where cos x = tan x to see the part of the interval that works.

Answer: (d) \( -\sqrt{1+\sin A}+\sqrt{1-\sin A} \)

Explanation:
1. \( \sqrt{1+\sin A}=\left|\sin\frac A2+\cos\frac A2\right| \) and \( \sqrt{1-\sin A}=\left|\sin\frac A2-\cos\frac A2\right| \) with \( \frac A2=170^\circ \).
2. At \( 170^\circ \): \( \sin\approx0.174,\ \cos\approx-0.985 \), so \( \sin+\cos\lt 0 \) and \( \sin-\cos\gt 0 \).
3. \( \sqrt{1+\sin A}=-(\sin\frac A2+\cos\frac A2) \) and \( \sqrt{1-\sin A}=\sin\frac A2-\cos\frac A2 \), so \( 2\sin\frac A2=\sqrt{1-\sin A}-\sqrt{1+\sin A} \).

Teacher's Note:
1. Decide the sign inside each modulus from the quadrant of A/2 (second quadrant here).
2. Add the two expressions to isolate 2 sin(A/2).

Answer: (d) natural number

Explanation:
1. \( \csc10^\circ-\sqrt3\sec10^\circ=\frac{\cos10^\circ-\sqrt3\sin10^\circ}{\sin10^\circ\cos10^\circ} \).
2. Numerator \( =2\cos(10^\circ+60^\circ)=2\cos70^\circ=2\sin20^\circ \); denominator \( =\frac12\sin20^\circ \).
3. The value is \( 4 \), a natural number.

Teacher's Note:
1. Write cos 10° − √3 sin 10° as 2 cos(10° + 60°).
2. Use sin 10° cos 10° = ½ sin 20°.

Answer: (b) 7

Explanation:
1. On \( C_k \) the particle covers an arc of k cm, i.e. an angle of \( \frac{k}{k}=1 \) radian.
2. After \( C_1,\dots,C_6 \) the total angle is 6 rad, which is less than \( 2\pi\approx6.283 \).
3. On \( C_7 \) the angle passes 6.283, so the particle first crosses the positive x-axis on \( C_7 \): n = 7.

Teacher's Note:
1. An arc of length k on a circle of radius k subtends exactly 1 radian.
2. Compare the cumulative angle with 2π.

Answer: (b) \( \frac\pi8 \) and \( \frac{3\pi}8 \)

Explanation:
1. If the hypotenuse is c and an acute angle is A, the altitude to the hypotenuse is \( h=c\sin A\cos A=\frac c2\sin2A \).
2. \( c=2\sqrt2h\Rightarrow h=\frac{c}{2\sqrt2}\Rightarrow\sin2A=\frac1{\sqrt2} \).
3. \( 2A=\frac\pi4 \) or \( \frac{3\pi}4 \), so the angles are \( \frac\pi8 \) and \( \frac{3\pi}8 \).

Teacher's Note:
1. The altitude to the hypotenuse is c sin A cos A.
2. The two acute angles add up to π/2.

Answer: (a) 1

Explanation:
1. Let \( x,y,z=\cot\frac A2,\cot\frac B2,\cot\frac C2 \). The expression is \( \frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}=\frac1{z^2}+\frac1{x^2}+\frac1{y^2} \).
2. That is \( \tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2 \).
3. With \( \tan\frac A2\tan\frac B2+\tan\frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1 \) and \( a^2+b^2+c^2\ge ab+bc+ca \), the minimum is 1, at the equilateral triangle.

Teacher's Note:
1. Cancel the common factors to turn the expression into a sum of tan² half-angles.
2. Use the triangle identity for the pairwise products of tan(A/2), tan(B/2), tan(C/2).

Answer: (b) an irrational number

Explanation:
1. \( \cot\theta-\tan\theta=2\cot2\theta \), so \( \cot7.5^\circ-\tan7.5^\circ=2\cot15^\circ=2(2+\sqrt3) \).
2. \( \tan67.5^\circ-\cot67.5^\circ=-2\cot135^\circ=2 \).
3. Sum \( =4+2\sqrt3+2=6+2\sqrt3=2(3+\sqrt3) \), an irrational number (and different from options (c), (d)).

Teacher's Note:
1. Use cot θ − tan θ = 2 cot 2θ for each pair.
2. cot 15° = 2 + √3.

Answer: (c) 11

Explanation:
1. The left side is \( \sum_{k=0}^{5}\cos2k\theta \), a cosine series in A.P. with 6 terms and common difference \( 2\theta \).
2. \( \sum_{k=0}^{5}\cos2k\theta=\frac{\sin(6\theta)\cos(5\theta)}{\sin\theta} \) (sum of n terms: \( \frac{\sin\frac{n d}{2}}{\sin\frac d2}\cos\left(a+\frac{(n-1)d}{2}\right) \) with \( a=0,d=2\theta,n=6 \)).
3. So \( m=5,\ n=6 \) and \( m+n=11 \).

Teacher's Note:
1. Use the formula for the sum of cosines in A.P.
2. Identify m and n by comparing with cos mθ sin nθ / sin θ.

Answer: (c) 12

Explanation:
1. Let \( t=x\sin x\gt 0 \). The expression is \( 9t+\frac4t \).
2. By AM-GM, \( 9t+\frac4t\ge2\sqrt{36}=12 \) with equality at \( t=\frac23 \), which is attainable because \( x\sin x \) takes every value between 0 and about 1.82 on \( (0,\pi) \).
3. Minimum value 12.

Teacher's Note:
1. Substitute t = x sin x and apply AM-GM.
2. Check that the equality value of t lies in the range of x sin x.

Answer: (b) B and D only

Explanation:
1. Pair terms symmetrically about \( 4x \): numerator \( =\cos4x\,(1+2\cos x+2\cos2x+2\cos3x) \) and denominator \( =\sin4x\,(1+2\cos x+2\cos2x+2\cos3x) \), so \( y=\cot4x \).
2. (A) \( y=\cot\frac\pi2=0 \), which is defined: false. (B) \( \cot\frac\pi4=1 \): true. (C) \( \cot\frac\pi8=\sqrt2+1\ne\sqrt2-1 \): false. (D) \( \cot\frac\pi{12}=2+\sqrt3 \): true.
3. So (B) and (D) hold.

Teacher's Note:
1. Use cos(4x − kx) + cos(4x + kx) = 2 cos 4x cos kx, and similarly for sines.
2. The common bracket cancels when it is non-zero.

Answer: (c) II, III and IV

Explanation:
1. \( \cos2\beta=1-2\sin^2\beta=1-2\sin\alpha\cos\alpha=1-\sin2\alpha \) (III is correct).
2. \( 2\sin^2\left(\frac\pi4-\alpha\right)=1-\cos\left(\frac\pi2-2\alpha\right)=1-\sin2\alpha \) (II is correct).
3. \( 2\cos^2\left(\frac\pi4+\alpha\right)=1+\cos\left(\frac\pi2+2\alpha\right)=1-\sin2\alpha \) (IV is correct); I is not.

Teacher's Note:
1. Start from cos 2β = 1 − 2 sin²β.
2. Rewrite the other options with the half-angle formulas.

Answer: (b) B and D only

Explanation:
1. Distances of the chords from the centre: \( R\cos\frac\theta2 \) and \( R\cos\theta \). On the same side, the distance between them is \( R\left(\cos\frac\theta2-\cos\theta\right) \).
2. With \( \cos\theta=2\cos^2\frac\theta2-1 \): \( R\left(1+\cos\frac\theta2-2\cos^2\frac\theta2\right)=R\left(1-\cos\frac\theta2\right)\left(1+2\cos\frac\theta2\right) \) (B).
3. Also \( \cos\frac\theta2-\cos\theta=2\sin\frac{3\theta}4\sin\frac\theta4 \) (D). Options (A) and (C) are not equal to this.

Teacher's Note:
1. A chord subtending angle φ at the centre is at distance R cos(φ/2) from the centre.
2. Express the difference of cosines as a product or in terms of cos(θ/2).