JEE Mathematics Trigonometric Functions MCQs Set 14

Answer: (b) 10

Explanation:
1. Let \(s=\sin^2\alpha\). Then \(\dfrac1{1-s}+\dfrac1{1+s}=\dfrac2{1-s^2}\).
2. \(\dfrac2{1-s^2}+\dfrac2{1+s^2}=\dfrac4{1-s^4}\) and \(\dfrac4{1-s^4}+\dfrac4{1+s^4}=\dfrac8{1-s^8}\).
3. \(s^8=\sin^{16}\alpha=\tfrac15\), so the value is \(\dfrac8{4/5}=10\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (c) \(2\sqrt{1+k}\)

Explanation:
1. With \(\theta=\arcsin k\): \(a=\theta,\ b=\pi-\theta,\ c=2\pi+\theta,\ d=3\pi-\theta\).
2. Half-angle sines: \(\sin\tfrac\theta2,\ \cos\tfrac\theta2,\ -\sin\tfrac\theta2,\ -\cos\tfrac\theta2\).
3. Sum = \(4s+3c-2s-c=2(s+c)=2\sqrt{1+\sin\theta}=2\sqrt{1+k}\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (d) \(\sqrt3\)

Explanation:
1. \(\tan^2\tfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}=\dfrac{(2-\cos\beta)-(2\cos\beta-1)}{(2-\cos\beta)+(2\cos\beta-1)}=\dfrac{3(1-\cos\beta)}{1+\cos\beta}\).
2. So \(\tan^2\tfrac\alpha2=3\tan^2\tfrac\beta2\); both tangents are positive, so the ratio is \(\sqrt3\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (a) \(\dfrac{\sqrt5+1}2\)

Explanation:
1. With \(5\alpha=\pi\): \(\sin7\alpha=-\sin2\alpha,\ \sin5\alpha=0,\ \sin3\alpha=\sin2\alpha,\ \sin6\alpha=-\sin\alpha,\ \sin4\alpha=\sin\alpha\).
2. The ratio is \(\dfrac{16\sin2\alpha+12\sin\alpha}{4\sin\alpha+12\sin2\alpha}=\dfrac{8\cos\alpha+3}{1+6\cos\alpha}\).
3. With \(\cos36^\circ=\dfrac{\sqrt5+1}4\) this simplifies to \(\dfrac{\sqrt5+1}2\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (c) \(\sqrt{\dfrac qp}\)

Explanation:
1. \(\cot x=\dfrac{1+\cos2x}{\sin2x}\) with \(2x=\tfrac\pi2+\theta\) gives \(\cot\!\left(\tfrac\pi4+\tfrac\theta2\right)=\dfrac{1-\sin\theta}{\cos\theta}=\sqrt{\dfrac{1-\sin\theta}{1+\sin\theta}}\).
2. \(\sin\theta=\dfrac{p-q}{p+q}\): \(1-\sin\theta=\dfrac{2q}{p+q}\) and \(1+\sin\theta=\dfrac{2p}{p+q}\), so the value is \(\sqrt{q/p}\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (b) \(\dfrac58\)

Explanation:
1. Square and add: \(2+2\cos(2\theta-2\phi)=\dfrac14+\dfrac94=\dfrac52\), so \(\cos(2\theta-2\phi)=\dfrac14\).
2. \(\cos^2(\theta-\phi)=\dfrac{1+\cos(2\theta-2\phi)}2=\dfrac58\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (d) \(4(1+\cos\theta)(1+\cos2\theta)\)

Explanation:
1. Let \(A=1+\cos\theta,\ B=1+\cos2\theta,\ s=\sin\theta,\ S=\sin2\theta\): \(u=AB-sS,\ v=sB+SA\).
2. \(u^2+v^2=(A^2+s^2)(B^2+S^2)\) since the cross terms cancel.
3. \(A^2+s^2=2+2\cos\theta\) and \(B^2+S^2=2+2\cos2\theta\), so \(u^2+v^2=4(1+\cos\theta)(1+\cos2\theta)\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (a) \(\dfrac1{16^{16}}\)

Explanation:
1. Let \(N=2^{64}-1\) and \(\theta=\dfrac{2\pi}N\). The product is \(\cos\theta\cos2\theta\cdots\cos2^{63}\theta=\dfrac{\sin2^{64}\theta}{2^{64}\sin\theta}\).
2. \(2^{64}\theta=\dfrac{2\pi(N+1)}N=2\pi+\theta\), so \(\sin2^{64}\theta=\sin\theta\).
3. The product is \(2^{-64}=16^{-16}\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (b) \(\dfrac{\sin x}{256\sin\dfrac x{256}}\)

Explanation:
1. Put \(\theta=\dfrac x{256}\); the product is \(\cos\theta\cos2\theta\cdots\cos128\theta\) (8 factors).
2. It equals \(\dfrac{\sin256\theta}{256\sin\theta}=\dfrac{\sin x}{256\sin(x/256)}\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (c) 1

Explanation:
1. \(3\sin x-4\cos x\le5\), so the argument is at most 20.
2. \(\log_{20}20=1\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (d) 3

Explanation:
1. Case 1, base = 1 in absolute value: \(2\sin x-\sqrt3=-1\) gives \(\sin x=\dfrac{\sqrt3-1}2\), two solutions in \( [0, \pi] \); \(2\sin x-\sqrt3=1\) needs \(\sin x>1\), impossible.
2. Case 2, exponent 0: \(2\cos^2x-3\cos x+1=0\) gives \(\cos x=1\) or \(\tfrac12\), i.e. \(x=0\) or \(\tfrac\pi3\). At \(x=\tfrac\pi3\) the base is 0 (\( 0^{0} \) undefined), so only \( x = 0 \) counts.
3. Total \( 2 + 1 = 3 \) solutions.

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (a) 4

Explanation:
1. \(\sin x-\sqrt3\cos x\) takes all values in \([-2,2]\), so we need \(|4m-6|\le2|4-m|\).
2. Squaring: \(3m^2-4m-7\le0\), i.e. \((3m-7)(m+1)\le0\), so \(-1\le m\le\tfrac73\).
3. Integers: \( -1, 0, 1, 2 \), giving 4 values.

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (d) 1

Explanation:
1. \(\tan\tfrac\theta2(1+\sec\theta)=\tan\theta\), and in general \(\tan t\,(1+\sec2t)=\tan2t\), so \(f_n(\theta)=\tan2^n\theta\).
2. \(f_4(\pi/64)=\tan\dfrac{16\pi}{64}=\tan\dfrac\pi4=1\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (c) \((a-b)^2\)

Explanation:
1. The product of the two radicands is \(a^2b^2+(a^2-b^2)^2\sin^2\theta\cos^2\theta\), so \(u^2=a^2+b^2+2\sqrt{a^2b^2+\tfrac14(a^2-b^2)^2\sin^22\theta}\).
2. Maximum \( (\sin 2 \theta = \pm 1) \): \(a^2+b^2+(a^2+b^2)=2(a^2+b^2)\). Minimum \( (\sin 2 \theta = 0) \): \(a^2+b^2+2ab=(a+b)^2\).
3. Difference = \(2a^2+2b^2-(a+b)^2=(a-b)^2\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (a) 0

Explanation:
1. Multiply by \(\sin x\cos x\): \(\sqrt3\sin x+\cos x+2(\sin^2x-\cos^2x)=0\), i.e. \(2\sin(x+\tfrac\pi6)=2\cos2x\).
2. \(\sin(x+\tfrac\pi6)=\sin(\tfrac\pi2-2x)\) gives \(x=\tfrac\pi9+\tfrac{2k\pi}3\) (giving \(\tfrac\pi9,\tfrac{7\pi}9,-\tfrac{5\pi}9\)) or \(x=-\tfrac\pi3\); all four lie in \( S \) and a numerical scan finds no others.
3. Their sum is \(\tfrac{-5-3+1+7}9\pi=0\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (b) \(\dfrac1{2^{n/2}}\)

Explanation:
1. \(\prod\cot\alpha_i=1\) means \(\prod\cos\alpha_i=\prod\sin\alpha_i\).
2. So \(\left(\prod\cos\alpha_i\right)^2=\prod\sin\alpha_i\cos\alpha_i=\prod\dfrac{\sin2\alpha_i}2\le\dfrac1{2^n}\), giving the maximum \(2^{-n/2}\) (when all angles are \(\pi/4\)).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (b) \(\dfrac{119+120\cot\alpha}{13}\)

Explanation:
1. \(\tan\beta<0\) and \(\sin\beta>0\) give \(\cos\beta=-\dfrac5{13}\).
2. The expression is \(\sin\alpha(5\cos\beta+12\sin\beta)+\cos\alpha(5\sin\beta-12\cos\beta)=\dfrac{119}{13}\sin\alpha+\dfrac{120}{13}\cos\alpha\).
3. Multiplying by \( \csc \alpha \) gives \( (119 + 120 \cot \alpha)/13 \).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (c) \(\dfrac{2k}{k-1}\)

Explanation:
1. \(X=\dfrac{\sin3A}{\sin A}=3-4\sin^2A\) and \(Y=\dfrac{\cos3A}{\cos A}=4\cos^2A-3=1-4\sin^2A\), so \(X-Y=2\).
2. \(\dfrac XY=k\): \(Y(k-1)=2\), so \(Y=\dfrac2{k-1}\) and \(X=\dfrac{2k}{k-1}\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (d) 0

Explanation:
1. Let the common value be \(t\): \(\dfrac1x=\dfrac{\cos\theta}t\), and so on.
2. \(\dfrac1x+\dfrac1y+\dfrac1z=\dfrac1t\left[\cos\theta+\cos(\theta+\tfrac{2\pi}3)+\cos(\theta+\tfrac{4\pi}3)\right]=0\), so \(xy+yz+zx=0\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (a) 6

Explanation:
1. \(\sin^3x=\dfrac{3\sin x-\sin3x}4\), so the left side is \(\dfrac{3\sin x\sin3x-\sin^23x}4\).
2. \(\sin x\sin3x=\tfrac12(\cos2x-\cos4x)\) and \(\sin^23x=\tfrac12(1-\cos6x)\), so the highest harmonic is \(\cos6x\): \( n = 6 \).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (c) \(3(m^2-1)^2\)

Explanation:
1. \(1-P_6=1-(\sin^6\theta+\cos^6\theta)=3\sin^2\theta\cos^2\theta\).
2. \(m^2=1+2\sin\theta\cos\theta\), so \(\sin\theta\cos\theta=\dfrac{m^2-1}2\).
3. \(4(1-P_6)=12\cdot\dfrac{(m^2-1)^2}4=3(m^2-1)^2\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (b) 8

Explanation:
1. Let \(s=\sin^2\theta,\ c=\cos^2\theta\) with \(s+c=1\): \(P_{10}=s^5+c^5,\ P_8=s^4+c^4,\ P_6=s^3+c^3\).
2. Substituting \(c=1-s\) makes the expression constant (checked: at \(\theta=0.7\) the value is 8.0), and at \(\theta=0\): \(6-15+10+7=8\).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (d) \(1-\dfrac{\sqrt{69}}6\)

Explanation:
1. First equation (replace 2 by \(2\sin^2x+2\cos^2x\)): \(3\tan^2\alpha+3\tan\alpha-5=0\), so \(\tan\alpha=\dfrac{-3\pm\sqrt{69}}6\).
2. Second equation: \(\sin^2x-\cos2x=2-\sin2x\) becomes \(2\sin^2x-\cos^2x=2(\sin^2x+\cos^2x)-2\sin x\cos x\), i.e. \(\cos x(2\sin x-3\cos x)=0\), so with \(\cos\beta\ne0\): \(\tan\beta=\tfrac32\).
3. \(\tan\alpha+\tan\beta=\dfrac{6\pm\sqrt{69}}6=1\pm\dfrac{\sqrt{69}}6\); the option given is \( 1 - \sqrt{69}/6 \).

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.

Answer: (a) 6

Explanation:
1. Using \( \sin t / \cos 3t = (\tan 3t - \tan t)/2 \) gives the sum as \( (\tan 27x - \tan x)/2 = 0 \) (provided the cosines are non-zero).
2. So \( \tan 27x = \tan x \), i.e. \( 26x = n \pi, x = n \pi / 26 \) in \( (0, \pi/4) \): \( n = 1, 2 \), ..., 6 (n pi/26 < \( \pi/4 \) gives \( n \) < 6.5). Each satisfies the cosine conditions, and a numerical scan also finds 6 sign changes.

Teacher's Note:
1. Reduce the expression with standard identities (sum-to-product, multiple angles) before substituting values.
2. Check the final answer numerically at a convenient angle when an identity is involved.