JEE Mathematics Trigonometric Functions MCQs Set 11

Answer: (c) 17

Explanation:
1. Adding the three angles gives \( \tan(\theta+2\theta+3\theta)=\frac{\sum\tan-\prod\tan}{1-\sum\tan\tan} \). The given equation makes the numerator zero, so \( \tan6\theta=0 \), i.e. \( \theta=\frac{k\pi}6 \).
2. The terms must be defined: \( \tan3\theta \) needs \( \theta\ne\frac\pi6+\frac{m\pi}3 \), which removes every odd k (\( \tan\theta \) and \( \tan2\theta \) impose nothing more). So k must be even.
3. \( k=2,4,\ldots,34 \) gives 17 values (k = 36 would be \( \theta=6\pi \), outside the open interval).

Teacher's Note:
1. The sum of the three tangents equals their product when the total angle is a multiple of pi.
2. Check which multiples of pi/6 keep all three tangents defined.

Answer: (c) 1.75

Explanation:
1. \( \sin^2x+\sin x\cos x=\frac{1-\cos2x}2+\frac{\sin2x}2=\frac12+\frac{\sin2x-\cos2x}2 \).
2. \( \sin2x-\cos2x\in[-\sqrt2,\sqrt2] \), so \( n\in\left[\frac{1-\sqrt2}2,\frac{1+\sqrt2}2\right] \).
3. \( p^3+q^3=2\left(\frac18+3\cdot\frac12\cdot\frac12\right)=\frac74=1.75 \) (using \( (a-b)^3+(a+b)^3=2a^3+6ab^2 \) with \( a=\frac12,\ b=\frac{\sqrt2}2 \)).

Teacher's Note:
1. Rewrite in terms of double angles to find the range of n.
2. The sum of two cubes with symmetric terms simplifies a lot.

Answer: (c) 6

Explanation:
1. Put \( t=\tan x \): \( \frac1t-\frac{4t}{1+t^2}=1 \Rightarrow 1+t^2-4t^2=t+t^3 \Rightarrow t^3+3t^2+t-1=0 \).
2. The cubic \( (t+1)(t^2+2t-1)=0 \) has three real roots: \( t=-1,\ -1\pm\sqrt2 \).
3. Each value of tan x gives two angles in (0, 2 pi), so there are 6 solutions.

Teacher's Note:
1. Express cot x and sin 2x in terms of tan x.
2. Check that the cubic factors as (t + 1)(t^2 + 2t - 1).

Answer: (a) 0

Explanation:
1. \( 3x^2+2x+3=3\left(x+\frac13\right)^2+\frac83\ge\frac83 \gt 2\ge2\sin x \).
2. The right side always exceeds the left side, so there is no real solution.

Teacher's Note:
1. Complete the square on the quadratic side.
2. Compare the minimum of the quadratic with the maximum of the sine.

Answer: (b) All four

Explanation:
1. \( \frac{\cos11^\circ+\sin11^\circ}{\cos11^\circ-\sin11^\circ}=\frac{1+\tan11^\circ}{1-\tan11^\circ}=\tan56^\circ \).
2. \( \tan304^\circ=\tan(304^\circ-180^\circ)=\tan124^\circ=-\tan56^\circ \), so \( -\tan304^\circ=\tan56^\circ \).
3. \( \cot214^\circ=\cot34^\circ=\tan56^\circ \). All four are equal.

Teacher's Note:
1. Divide numerator and denominator by cos 11 to get the tangent formula.
2. Reduce each angle to the first quadrant.

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

Explanation:
1. With \( m=\tan\frac t2 \): \( \frac{2m}{1+m^2}+\frac{1-m^2}{1+m^2}=\frac15 \Rightarrow 5(2m+1-m^2)=1+m^2 \Rightarrow 3m^2-5m-2=0 \).
2. \( (3m+1)(m-2)=0 \), so \( m=2 \) or \( -\frac13 \).

Teacher's Note:
1. Use the half-angle tangent substitution.
2. Solve the resulting quadratic.

Answer: (d) (ii) and (iv) only

Explanation:
1. \( \frac{\sin x+\cos x}{\cos^3x}=\frac{\sin x}{\cos^3x}+\frac1{\cos^2x}=\tan x\sec^2x+\sec^2x=(1+\tan x)\sec^2x \).
2. With \( t=\tan x \): \( (1+t)(1+t^2)=1+t+t^2+t^3 \). So (ii) and (iv) are correct.

Teacher's Note:
1. Split the fraction and use sec^2 = 1 + tan^2.
2. Expand (1 + t)(1 + t^2).

Answer: (a) (ii) and (iii) only

Explanation:
1. 1 radian is about \( 57.3^\circ \), which is larger than \( 1^\circ \), and both lie in \( (0,\frac\pi2) \).
2. Sine increases and cosine decreases there, so \( \sin1^\circ \lt \sin1 \) and \( \cos1^\circ \gt \cos1 \).

Teacher's Note:
1. The unit matters: 1 degree is much smaller than 1 radian.
2. Use monotonicity on the first quadrant.

Answer: (c) (i), (ii) and (iii)

Explanation:
1. From \( \sin x+\sin y=a,\ \cos x+\cos y=b \): \( \tan\frac{x+y}2=\frac ab \) and \( a^2+b^2=4\cos^2\frac{x-y}2 \).
2. \( \sin(x+y)=\frac{2\tan\frac{x+y}2}{1+\tan^2\frac{x+y}2}=\frac{2ab}{a^2+b^2} \) and \( \cos(x+y)=\frac{b^2-a^2}{a^2+b^2} \), so (iv) is false.
3. \( \tan^2\frac{x-y}2=\sec^2\frac{x-y}2-1=\frac{4}{a^2+b^2}-1 \), so tan((x-y)/2) can have either sign: (ii) and (iii) can be true.

Teacher's Note:
1. Use sum-to-product on both given equations and divide.
2. The sign of tan((x - y)/2) depends on whether x > y or x < y.

Answer: (d) (ii) and (iii) only

Explanation:
1. \( \tan A\tan B=2 \Rightarrow \sin A\sin B=2\cos A\cos B \).
2. \( \cos(A-B)=\cos A\cos B+\sin A\sin B=3\cos A\cos B=\frac35 \), so \( \cos A\cos B=\frac15 \) and \( \sin A\sin B=\frac25 \).
3. \( \cos(A+B)=\frac15-\frac25=-\frac15 \). The value of \( \sin A\cos B \) is not fixed by the data.

Teacher's Note:
1. Turn tan A tan B = 2 into a relation between the products of sines and cosines.
2. cos(A - B) is the sum of the two products.

Answer: (a) (i) and (iii) only

Explanation:
1. The legs are \( 2\cos(\alpha+\beta)\left[1+\cos(\alpha-\beta)\right] \) and \( 2\sin(\alpha+\beta)\left[1+\cos(\alpha-\beta)\right] \), using \( \cos2\alpha+\cos2\beta=2\cos(\alpha+\beta)\cos(\alpha-\beta) \) and the sine analogue.
2. Hypotenuse \( =2\left[1+\cos(\alpha-\beta)\right]\sqrt{\cos^2(\alpha+\beta)+\sin^2(\alpha+\beta)}=2\left[1+\cos(\alpha-\beta)\right]=4\cos^2\frac{\alpha-\beta}2 \).
3. Options (ii) and (iv) involve \( \alpha+\beta \) instead of \( \alpha-\beta \); for \( \alpha=\beta=0 \) the hypotenuse is 4 but \( 2[1-\cos0]=0 \), so they are false.

Teacher's Note:
1. Factor the common term 1 + cos(alpha - beta) out of both legs.
2. Test the rejected options with alpha = beta = 0.

Answer: (b) (ii) and (iv) only

Explanation:
1. With \( t=\tan\frac\alpha2 \): \( 2(a+2)t+(2a-1)(1-t^2)=(2a+1)(1+t^2) \Rightarrow 2at^2-(a+2)t+1=0 \Rightarrow (2t-1)(at-1)=0 \).
2. \( t=\frac12 \Rightarrow \tan\alpha=\frac{2t}{1-t^2}=\frac43 \). \( t=\frac1a \Rightarrow \tan\alpha=\frac{2a}{a^2-1} \).

Teacher's Note:
1. Use the half-angle tangent substitution for the linear equation in sin and cos.
2. Each root of the quadratic in t gives one value of tan alpha.

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

Explanation:
1. \( y+z=a(\cos^2x+\sin^2x)+c(\sin^2x+\cos^2x)=a+c \).
2. \( y-z=(a-c)\cos2x+2b\sin2x \). With \( \tan x=\frac{2b}{a-c} \): \( \cos2x=\frac{(a-c)^2-4b^2}{(a-c)^2+4b^2} \) and \( \sin2x=\frac{4b(a-c)}{(a-c)^2+4b^2} \).
3. \( y-z=\frac{(a-c)[(a-c)^2-4b^2]+8b^2(a-c)}{(a-c)^2+4b^2}=a-c \).

Teacher's Note:
1. Use the double-angle formulas in terms of tan x.
2. Simplify the numerator to (a - c)[(a - c)^2 + 4b^2].

Answer: (b) (ii) and (iv) only

Explanation:
1. \( \sin^6x+\cos^6x=1-\frac34\sin^22x\in\left[\frac14,1\right] \).
2. So \( a^2\in\left[\frac14,1\right] \), i.e. \( \frac12\le|a|\le1 \). The intervals in (ii) and (iv) lie inside this set; (i) and (iii) contain values such as 0 that are not allowed.

Teacher's Note:
1. Find the range of sin^6 x + cos^6 x.
2. The square root of the range gives the allowed |a|.

Answer: (b) (ii) and (iv) only

Explanation:
1. \( \sin(x-y)=\frac12 \) with \( x-y\in(-\pi,\pi) \): \( x-y=\frac\pi6 \) or \( \frac{5\pi}6 \). \( \cos(x+y)=\frac12 \) with \( x+y\in(0,2\pi) \): \( x+y=\frac\pi3 \) or \( \frac{5\pi}3 \).
2. Combining: \( (x-y,x+y)=(\frac\pi6,\frac\pi3) \Rightarrow x=\frac\pi4,\ y=\frac\pi{12} \); \( (\frac\pi6,\frac{5\pi}3) \Rightarrow x=\frac{11\pi}{12},\ y=\frac{3\pi}4 \); the other two combinations give y < 0 or x > pi.

Teacher's Note:
1. List the possible values of each combination within the allowed ranges.
2. Check that x and y both lie in (0, pi).

Answer: (d) (i) and (iv) only

Explanation:
1. Put \( t=\tan^2\alpha \): \( \sec^2\alpha=1+t,\ \operatorname{cosec}^2\alpha=1+\frac1t \).
2. The left side becomes \( \frac1{t^2}-t^2 \). Setting \( \frac1{t^2}-t^2=\frac{15}4 \) and writing \( s=t^2 \): \( 4s^2+15s-4=0 \Rightarrow s=\frac14 \).
3. So \( \tan^4\alpha=\frac14 \), \( \tan^2\alpha=\frac12 \) and \( \tan\alpha=\pm\frac1{\sqrt2} \).

Teacher's Note:
1. Write everything in terms of tan^2 alpha.
2. Only the positive root of the quadratic in tan^4 alpha is acceptable.

Answer: (d) \( \frac{1+q}q \)

Explanation:
1. Multiply the sum by \( q=\cos A\cos B\cos C \): \( \sum\sin A\sin B\cos C \).
2. Since \( \sin A\sin B=\cos C+\cos A\cos B \), \( \sum\sin A\sin B\cos C=\sum\cos^2C+3q=(1-2q)+3q=1+q \).
3. So the required sum is \( \frac{1+q}q \).

Teacher's Note:
1. In a triangle cos^2 A + cos^2 B + cos^2 C = 1 - 2 cos A cos B cos C.
2. Use cos C = sin A sin B - cos A cos B.

Answer: (a) \( \frac{p^3-3pq}{q^3} \)

Explanation:
1. Let \( e_1=\sum\tan A=\frac pq \), \( e_2=\sum\tan A\tan B=\frac{1+q}q \), \( e_3=\tan A\tan B\tan C=\frac pq \).
2. \( \sum\tan^3A=e_1^3-3e_1e_2+3e_3=\frac{p^3}{q^3}-\frac{3p(1+q)}{q^2}+\frac{3p}q \).
3. \( =\frac{p^3}{q^3}-\frac{3p}{q^2}=\frac{p^3-3pq}{q^3} \).

Teacher's Note:
1. Use the identity for the sum of cubes in terms of elementary symmetric sums.
2. The terms 3p/q cancel.

Answer: (d) six real roots

Explanation:
1. \( 4s^3+2s^2-2s-1=(2s+1)(2s^2-1) \), so \( \sin x=-\frac12 \) or \( \sin x=\pm\frac1{\sqrt2} \).
2. \( \sin x=-\frac12 \): \( x=-\frac\pi6,-\frac{5\pi}6 \). \( \sin x=\frac1{\sqrt2} \): \( x=\frac\pi4,\frac{3\pi}4 \). \( \sin x=-\frac1{\sqrt2} \): \( x=-\frac\pi4,-\frac{3\pi}4 \).
3. Six roots in all.

Teacher's Note:
1. Factor the cubic by grouping.
2. Each value of sin x in (-1, 1) has two solutions in [-pi, pi] here.

Answer: (a) one solution

Explanation:
1. If \( \tan x\ge0 \) the equation gives \( 0=\sec x \), impossible.
2. If \( \tan x \lt 0 \): \( -\tan x=\tan x+\sec x \Rightarrow -2\sin x=1 \Rightarrow \sin x=-\frac12 \).
3. With \( \tan x \lt 0 \) we need \( \cos x \gt 0 \) (sin is negative), so \( x=\frac{11\pi}6 \) only.

Teacher's Note:
1. Split into the cases tan x >= 0 and tan x < 0.
2. Check the sign of cos x for sin x = -1/2.

Answer: (c) \( \frac{n\pi}2+\frac\pi6 \lt x \lt \frac{n\pi}2+\frac\pi3 \)

Explanation:
1. \( \sin^6x+\cos^6x=1-\frac34\sin^22x \lt \frac7{16} \Rightarrow \sin^22x \gt \frac34 \).
2. \( |\sin2x| \gt \frac{\sqrt3}2 \Rightarrow 2x\in\left(k\pi+\frac\pi3,k\pi+\frac{2\pi}3\right) \).
3. \( x\in\left(\frac{k\pi}2+\frac\pi6,\frac{k\pi}2+\frac\pi3\right) \).

Teacher's Note:
1. Reduce to a condition on |sin 2x|.
2. |sin t| > c repeats with period pi.

Answer: (b) \( \left[-\frac{2\pi}3,\frac{2\pi}3\right] \)

Explanation:
1. \( 2\cos^2x+5\cos x+2\ge0 \Rightarrow (2\cos x+1)(\cos x+2)\ge0 \).
2. \( \cos x+2 \gt 0 \), so we need \( \cos x\ge-\frac12 \), giving \( x\in\left[-\frac{2\pi}3,\frac{2\pi}3\right] \) in \( [-\pi,\pi] \).

Teacher's Note:
1. Factor the quadratic in cos x.
2. cos x + 2 is always positive.

Answer: (a) 2

Explanation:
1. The denominator \( =\frac{1-\cos2\theta}2+\frac32\sin2\theta+\frac{5(1+\cos2\theta)}2=3+2\cos2\theta+\frac32\sin2\theta \).
2. Its minimum is \( 3-\sqrt{4+\frac94}=3-\frac52=\frac12 \), so the maximum of the reciprocal is 2.

Teacher's Note:
1. Write the denominator as a constant plus a cos 2 theta + b sin 2 theta.
2. The reciprocal is largest when the denominator is smallest.

Answer: (d) 7

Explanation:
1. Put \( \theta=\frac\pi n \): \( \frac1{\sin\theta}-\frac1{\sin3\theta}=\frac1{\sin2\theta} \Rightarrow \frac{\sin3\theta-\sin\theta}{\sin\theta\sin3\theta}=\frac1{\sin2\theta} \).
2. \( \sin3\theta-\sin\theta=2\cos2\theta\sin\theta \), so \( 2\cos2\theta\sin2\theta=\sin3\theta \Rightarrow \sin4\theta=\sin3\theta \).
3. \( 4\theta=\pi-3\theta \Rightarrow \theta=\frac\pi7 \), i.e. \( n=7 \) (the other solution \( \theta=2k\pi \) is rejected).

Teacher's Note:
1. Bring the terms with sin theta and sin 3 theta together.
2. sin 4 theta = sin 3 theta gives 7 theta = pi.