JEE Mathematics Trigonometric Functions MCQs Set 08

Answer: (a) \( 0 \)

Explanation:
1. \( \cos(540^\circ-\theta)=\cos(180^\circ-\theta)=-\cos\theta \).
2. \( \sin(630^\circ-\theta)=\sin(270^\circ-\theta)=-\cos\theta \).
3. Difference: \( -\cos\theta+\cos\theta=0 \).

Teacher's Note:
1. Reduce multiples of 90 degrees first, then use the quadrant rule.
2. An odd multiple of 90 degrees swaps sine and cosine; an even multiple keeps the function.

Answer: (d) 0

Explanation:
1. \( \cos\frac{2\pi}3=\cos\frac{4\pi}3=-\frac12 \), so \( x=-\frac y2=-\frac z2 \), i.e. \( y=z=-2x \).
2. \( xy+yz+zx=-2x^2+4x^2-2x^2=0 \).

Teacher's Note:
1. Express everything in terms of x.
2. Both cosines are equal to minus one half.

Answer: (a) -1

Explanation:
1. \( \sin x=\frac1{\sqrt{10}},\ \tan x=\frac13 \).
2. The sum of logs is \( \log_{10}(\sin x\cos x\tan x)=\log_{10}(\sin^2x)=\log_{10}\frac1{10}=-1 \).

Teacher's Note:
1. Since \( \cos x\tan x=\sin x \), the product is just \( \sin^2x \).
2. Combine the logarithms before evaluating.

Answer: (a) \( a^4-4a^2+2 \)

Explanation:
1. \( \tan^2\alpha+\cot^2\alpha=a^2-2 \).
2. \( \tan^4\alpha+\cot^4\alpha=(a^2-2)^2-2=a^4-4a^2+2 \).

Teacher's Note:
1. Square twice, subtracting twice the product (which is 1) each time.
2. tan times cot is always 1.

Answer: (a) \( \frac12\left(\frac1\alpha-\alpha\right) \)

Explanation:
1. \( (\operatorname{cosec}\theta-\cot\theta)(\operatorname{cosec}\theta+\cot\theta)=1 \), so \( \operatorname{cosec}\theta+\cot\theta=\frac1\alpha \).
2. Subtract: \( 2\cot\theta=\frac1\alpha-\alpha \).

Teacher's Note:
1. The pair cosec + cot and cosec - cot are reciprocals.
2. Add or subtract the two to isolate cot.

Answer: (d) 25

Explanation:
1. Square and add: \( (a\cos\theta+b\sin\theta)^2+(a\sin\theta-b\cos\theta)^2=a^2+b^2 \).
2. So a^2 + b^2 = 9 + 16 = 25.

Teacher's Note:
1. The cross terms cancel when the two squares are added.
2. This is the standard Lagrange-type identity.

Answer: (c) \( \sin^2x \)

Explanation:
1. \( \tan(x-\frac\pi2)=-\cot x,\ \cos(\frac{3\pi}2+x)=\sin x,\ \sin(\frac{7\pi}2-x)=-\cos x \).
2. Numerator: \( -\cos x+\cos^3x=-\cos x\sin^2x \).
3. Denominator: \( \cos(x-\frac\pi2)=\sin x,\ \tan(\frac{3\pi}2+x)=-\cot x \), product \( -\cos x \).
4. Ratio \( =\sin^2x \).

Teacher's Note:
1. Reduce each angle to x plus a multiple of 90 degrees and apply the quadrant sign.
2. Odd multiples of a right angle convert tan to cot and sin to cos.

Answer: (c) \( \frac1{16} \)

Explanation:
1. \( \cos\frac{9\pi}{10}=-\cos\frac\pi{10} \) and \( \cos\frac{7\pi}{10}=-\cos\frac{3\pi}{10} \).
2. The product becomes \( \sin^2\frac\pi{10}\sin^2\frac{3\pi}{10}=(\sin18^\circ\sin54^\circ)^2 \).
3. \( \sin18^\circ\sin54^\circ=\sin18^\circ\cos36^\circ=\frac14 \), so the value is \( \frac1{16} \).

Teacher's Note:
1. Pair the factors with supplementary angles to form 1 - cos^2.
2. sin 18 cos 36 = 1/4 is a standard value.

Answer: (b) -1

Explanation:
1. \( \sin66^\circ=\cos24^\circ \): numerator \( =\sin24^\circ\cos6^\circ-\cos24^\circ\sin6^\circ=\sin18^\circ \).
2. \( \cos51^\circ=\sin39^\circ,\ \sin69^\circ=\cos21^\circ \): denominator \( =\sin21^\circ\cos39^\circ-\sin39^\circ\cos21^\circ=\sin(-18^\circ) \).
3. The ratio is -1.

Teacher's Note:
1. Convert the cosines and sines of complementary angles so that a compound-angle formula appears.
2. Watch the sign of the denominator.

Answer: (b) \( \frac{a^2}{a^2+(1-b)^2} \)

Explanation:
1. \( \tan A+\tan B=a,\ \tan A\tan B=b \), so \( \tan(A+B)=\frac a{1-b} \).
2. \( \sin^2(A+B)=\frac{\tan^2}{1+\tan^2}=\frac{a^2}{(1-b)^2+a^2} \).

Teacher's Note:
1. Use \( \sin^2\theta=\tan^2\theta/(1+\tan^2\theta) \).
2. Option (c) contains an undefined letter c.

Answer: (b) \( \frac{1-x^2}{2x} \)

Explanation:
1. \( \tan155^\circ=-\tan25^\circ=-x,\ \tan115^\circ=-\cot25^\circ=-\frac1x \).
2. The expression is \( \tan(155^\circ-115^\circ)=\tan40^\circ \) \( =\frac{-x+1/x}{1+1}=\frac{1-x^2}{2x} \).

Teacher's Note:
1. The expression is the tangent difference formula, but the answer must be in terms of x.
2. Substitute tan 155 and tan 115 in terms of tan 25.

Answer: (c) \( \frac12 \)

Explanation:
1. \( \frac{\cot A}{1+\cot A}=\frac1{1+\tan A} \), so the product is \( \frac1{(1+\tan A)(1+\tan B)} \).
2. \( \tan(A+B)=1 \Rightarrow \tan A+\tan B=1-\tan A\tan B \).
3. \( (1+\tan A)(1+\tan B)=1+(1-\tan A\tan B)+\tan A\tan B=2 \). The value is \( \frac12 \).

Teacher's Note:
1. Invert the cot expression to a tan expression.
2. If A + B = 45 degrees (mod 180) then (1 + tan A)(1 + tan B) = 2.

Answer: (d) 1

Explanation:
1. \( \tan203^\circ=\tan23^\circ \).
2. \( \tan(23^\circ+22^\circ)=1 \Rightarrow \tan23^\circ+\tan22^\circ=1-\tan23^\circ\tan22^\circ \).
3. Hence the required sum is 1.

Teacher's Note:
1. Reduce 203 degrees by subtracting 180.
2. 23 + 22 = 45 gives a ready-made relation.

Answer: (b) 0

Explanation:
1. \( \tan A=\frac43 \) in the third quadrant: \( \sin A=-\frac45,\ \cos A=-\frac35 \).
2. \( 5\sin2A=10\sin A\cos A=\frac{24}5 \), \( 3\sin A=-\frac{12}5 \), \( 4\cos A=-\frac{12}5 \).
3. Total \( \frac{24}5-\frac{12}5-\frac{12}5=0 \).

Teacher's Note:
1. Both sine and cosine are negative in the third quadrant.
2. Compute each term separately and add.

Answer: (c) 3

Explanation:
1. \( 2\sin\frac A2\sin\frac{5A}2=\cos2A-\cos3A \), so the expression is \( 8(1+\cos A)-16(\cos2A-\cos3A) \).
2. With \( c=\frac34 \): \( \cos2A=2c^2-1=\frac18,\ \cos3A=4c^3-3c=-\frac9{16} \).
3. \( 8\cdot\frac74-16\left(\frac18+\frac9{16}\right)=14-11=3 \).

Teacher's Note:
1. Convert the product of sines into a difference of cosines.
2. Then use the multiple-angle formulas for cos 2A and cos 3A.

Answer: (a) independent of \( \phi \)

Explanation:
1. \( \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{-2\tan^2\phi}{2+2\tan^2\phi}=-\sin^2\phi \).
2. So \( \cos2\theta+\sin^2\phi=0 \) for every \( \phi \).

Teacher's Note:
1. Put the given relation into the cos 2 theta formula.
2. The result is a constant, so no phi appears.

Answer: (b) \( \frac{4\sqrt3}3 \)

Explanation:
1. \( \cos290^\circ=\sin20^\circ,\ \sin250^\circ=-\cos20^\circ \).
2. Sum \( =\frac1{\sin20^\circ}-\frac1{\sqrt3\cos20^\circ}=\frac{\sqrt3\cos20^\circ-\sin20^\circ}{\sqrt3\sin20^\circ\cos20^\circ} \).
3. Numerator \( =2\cos50^\circ=2\sin40^\circ \) and denominator \( =\frac{\sqrt3}2\sin40^\circ \), so the value is \( \frac4{\sqrt3}=\frac{4\sqrt3}3 \).

Teacher's Note:
1. \( \sqrt3\cos20^\circ-\sin20^\circ=2\cos(20^\circ+30^\circ)=2\cos50^\circ=2\sin40^\circ \) uses the auxiliary-angle method.
2. Always sanity-check the final value on a calculator (about 2.309).

Answer: (a) \( \tan3A\tan2A\tan A \)

Explanation:
1. \( \tan3A=\tan(2A+A)=\frac{\tan2A+\tan A}{1-\tan2A\tan A} \).
2. So \( \tan3A-\tan3A\tan2A\tan A=\tan2A+\tan A \), i.e. \( \tan3A-\tan2A-\tan A=\tan3A\tan2A\tan A \).

Teacher's Note:
1. Expand tan 3A as tan(2A + A) and clear the denominator.
2. This identity is the basis of many conditional identities.

Answer: (c) 2

Explanation:
1. \( \sin70^\circ\sin50^\circ\sin10^\circ=\cos20^\circ\cos40^\circ\cos80^\circ=\frac18 \), so \( 8\sin70^\circ\sin50^\circ\sin10^\circ=1 \).
2. Numerator \( =\cos20^\circ+1=2\cos^210^\circ=2\sin^280^\circ \).
3. Hence the value is 2.

Teacher's Note:
1. \( \cos20^\circ\cos40^\circ\cos80^\circ=\frac18 \) is the standard chain of angles doubling.
2. Use \( 1+\cos20^\circ=2\cos^210^\circ=2\sin^280^\circ \).

Answer: (d) \( A=B \)

Explanation:
1. \( \tan6^\circ\tan66^\circ\tan54^\circ=\tan18^\circ \) and \( \tan42^\circ\tan78^\circ\tan18^\circ=\tan54^\circ \) (from \( \tan\theta\tan(60^\circ-\theta)\tan(60^\circ+\theta)=\tan3\theta \)); multiplying gives \( \tan6^\circ\tan42^\circ\tan66^\circ\tan78^\circ=1 \).
2. So \( A=\frac1{\tan66^\circ\tan78^\circ}=\cot66^\circ\cot78^\circ=B \).

Teacher's Note:
1. Use the identity tan t tan(60 - t) tan(60 + t) = tan 3t twice.
2. The product of the four tangents is 1, so A is the reciprocal of tan 66 tan 78.

Answer: (a) 1, 2, 3

Explanation:
1. In a triangle \( \tan A+\tan B+\tan C=\tan A\tan B\tan C=6 \), so \( \tan C=3 \).
2. Then \( \tan A+\tan B=3 \) and \( \tan A\tan B=2 \), so they are 1 and 2.
3. With \( \tan A\tan B=2 \) and the order asked, \( (\tan A,\tan B,\tan C)=(1,2,3) \).

Teacher's Note:
1. The sum of tangents equals their product when A + B + C = 180 degrees.
2. Solve the resulting quadratic.

Answer: (c) \( \cot\alpha \)

Explanation:
1. Use \( \tan\theta=\cot\theta-2\cot2\theta \).
2. \( \tan\alpha+2\tan2\alpha+4\tan4\alpha=(\cot\alpha-2\cot2\alpha)+2(\cot2\alpha-2\cot4\alpha)+4(\cot4\alpha-2\cot8\alpha)=\cot\alpha-8\cot8\alpha \).
3. Adding \( 8\cot8\alpha \) leaves \( \cot\alpha \).

Teacher's Note:
1. The identity cot t - 2 cot 2t = tan t telescopes.
2. Each term cancels with the next.

Answer: (d) \( -\frac{\cos(\pi/10)}{16} \)

Explanation:
1. Let \( \theta=\frac\pi{10} \). The product of \( \cos\theta\cos2\theta\cos4\theta\cos8\theta\cos16\theta=\frac{\sin32\theta}{32\sin\theta} \).
2. \( \sin32\theta=\sin\frac{16\pi}5=\sin\left(3\pi+\frac\pi5\right)=-\sin\frac\pi5=-\sin2\theta \).
3. So the product is \( -\frac{\sin2\theta}{32\sin\theta}=-\frac{2\sin\theta\cos\theta}{32\sin\theta}=-\frac{\cos(\pi/10)}{16} \).

Teacher's Note:
1. Multiply and divide by sin of the first angle to use the doubling formula repeatedly.
2. sin(16 pi / 5) is negative because 16 pi / 5 is just past 3 pi.

Answer: (d) \( \frac12 \)

Explanation:
1. There are 9 terms. Using \( \sum_{k=1}^{n}\cos(2k-1)x=\frac{\sin2nx}{2\sin x} \) with \( x=\frac\pi{19},\ n=9 \): \( \frac{\sin\frac{18\pi}{19}}{2\sin\frac\pi{19}} \).
2. \( \sin\frac{18\pi}{19}=\sin\frac\pi{19} \), so the sum is \( \frac12 \).

Teacher's Note:
1. Use the sum of cosines in arithmetic progression.
2. The final sine ratio is 1.