JEE Mathematics Trigonometric Functions MCQs Set 07

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

Explanation:
1. Since \( A+B<\frac\pi2 \): \( \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}>0 \) with \( \tan A+\tan B>0 \), so \( \tan A\tan B<1 \).
2. Also \( \cos A+\cos B+\cos C=1+\frac rR>1 \) for every triangle.
3. So (ii) and (iii) are true.

Teacher's Note:
1. In any triangle the sum of cosines is 1 + r/R, hence greater than 1.
2. If A + B is acute, the product of tangents is below 1.

Answer: (b) \( -\frac13 \)

Explanation:
1. \( \cos A+\cos B=2\cos\frac{A+B}2\cos\frac{A-B}2=2\cdot\frac{\sqrt3}2\cos\frac{A-B}2=1 \).
2. So \( \cos\frac{A-B}2=\frac1{\sqrt3} \) and \( \cos(A-B)=2\cdot\frac13-1=-\frac13 \).

Teacher's Note:
1. Use the sum-to-product formula and then the double-angle formula.
2. cos 2t = 2cos^2 t - 1.

Answer: (d) 4

Explanation:
1. Cosine rule: \( 8=16+a^2-4\sqrt3a \), so \( a^2-4\sqrt3a+8=0 \) with \( a=2\sqrt3\pm2 \).
2. Area \( =\frac12\cdot4\cdot a\sin30^\circ=a \), so the difference is \( (2\sqrt3+2)-(2\sqrt3-2)=4 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 1 : 2 : 3

Explanation:
1. Since \( 1^2+(\sqrt3)^2=2^2 \), the triangle is right-angled, with the right angle opposite the side 2.
2. The side 1 is opposite \( 30^\circ \) and \( \sqrt3 \) is opposite \( 60^\circ \).
3. Angles \( 30^\circ,60^\circ,90^\circ \) are in the ratio 1 : 2 : 3.

Teacher's Note:
1. Check the Pythagorean relation first.
2. A 30-60-90 triangle has sides 1 : root 3 : 2.

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

Explanation:
1. With \( s=\sin^2x \): \( \frac{s^2}2+\frac{(1-s)^2}3=\frac15 \) gives \( 25s^2-20s+4=0 \), so \( s=\frac25 \), \( \cos^2x=\frac35 \) and \( \tan^2x=\frac23 \).
2. Then \( \frac{(2/5)^4}{8}+\frac{(3/5)^4}{27}=\frac2{625}+\frac3{625}=\frac1{125} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \tan\beta+2\tan\gamma \)

Explanation:
1. \( \tan\alpha=\cot\beta \) and also \( \tan\alpha=\tan(\beta+\gamma)=\frac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma} \).
2. So \( \cot\beta(1-\tan\beta\tan\gamma)=\tan\beta+\tan\gamma \), i.e. \( \cot\beta-\tan\gamma=\tan\beta+\tan\gamma \).
3. Hence \( \tan\alpha=\cot\beta=\tan\beta+2\tan\gamma \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 1

Explanation:
1. The determinant equals \( (\sin x-\cos x)^2(\sin x+2\cos x) \).
2. In the interval \( \sin x=\cos x \) only at \( x=\frac\pi4 \), and \( \tan x=-2 \) is impossible there.
3. So there is exactly one root.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( 1+\sqrt3 \)

Explanation:
1. Cosine rule at \( C=\frac\pi6 \): \( (2x+1)^2=a^2+b^2-\sqrt3ab \), whose real roots are \( x=\pm1,\ 1+\sqrt3,\ -2-\sqrt3 \).
2. Sides must be positive: \( b>0 \) needs \( |x|>1 \) and \( c>0 \) needs \( x>-\frac12 \), so \( x=1+\sqrt3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. \( \frac{A-B+C}{2}=\frac{\pi-2B}{2}=\frac\pi2-B \).
2. So \( 2ac\sin\left(\frac\pi2-B\right)=2ac\cos B=a^2+c^2-b^2 \) by the cosine rule.

Teacher's Note:
1. Use A + B + C = pi to simplify the angle.
2. 2ac cos B = a squared + c squared - b squared.

Answer: (d) 3

Explanation:
1. Multiply the second equation by \( yz \): \( xyz\sin3\theta=2z\cos3\theta+2y\sin3\theta \).
2. Equating with the first and third equations gives \( zc+ys=0 \) and \( y(c-2s)=zc \) (with \( c=\cos3\theta,\ s=\sin3\theta \)); hence \( c=s \), i.e. \( \tan3\theta=1 \) (then \( y=-z \), \( x=0 \) works).
3. In \( (0,\pi) \): \( 3\theta=\frac\pi4,\frac{5\pi}4,\frac{9\pi}4 \): 3 values.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( 12+7\sqrt3 \)

Explanation:
1. The equal sides are \( b \) and the base is \( \sqrt3\,b \) (base angles \( 30^\circ \)).
2. Area \( =\frac12b^2\sin120^\circ=\frac{\sqrt3}{4}b^2 \); semi-perimeter \( s=\frac{(2+\sqrt3)b}{2} \).
3. From \( r=\frac{\text{Area}}{s}=\frac{\sqrt3b}{2(2+\sqrt3)}=\sqrt3 \): \( b=2(2+\sqrt3) \).
4. Area \( =\frac{\sqrt3}{4}\cdot4(7+4\sqrt3)=7\sqrt3+12 \).

Teacher's Note:
1. Use r = Area / s.
2. Express all lengths in terms of the equal side.

Answer: (b) \( x\sqrt{1+x^2} \)

Explanation:
1. With \( \theta=\cot^{-1}x \): \( \cos\theta=\frac{x}{\sqrt{1+x^2}} \), \( \sin\theta=\frac{1}{\sqrt{1+x^2}} \).
2. \( x\cos\theta+\sin\theta=\frac{x^2+1}{\sqrt{1+x^2}}=\sqrt{1+x^2} \), so the bracket is \( (1+x^2)-1=x^2 \) and its root is \( x \).
3. Result \( x\sqrt{1+x^2} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \sqrt3 \)

Explanation:
1. In AP: \( B=60^\circ \). By the sine rule \( \frac ac=\frac{\sin A}{\sin C} \).
2. The expression is \( 2\sin A\cos C+2\cos A\sin C=2\sin(A+C)=2\sin B=\sqrt3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \frac\pi{12} \) and \( \frac{5\pi}{12} \)

Explanation:
1. \( \operatorname{cosec}A\operatorname{cosec}(A+\tfrac\pi4)=\sqrt2[\cot A-\cot(A+\tfrac\pi4)] \), so the sum telescopes to \( \sqrt2[\cot\theta-\cot(\theta+\tfrac{3\pi}2)]=\sqrt2(\cot\theta+\tan\theta) \).
2. Set \( \sqrt2\cdot\frac2{\sin2\theta}=4\sqrt2 \): \( \sin2\theta=\frac12 \), so \( \theta=\frac\pi{12} \) or \( \frac{5\pi}{12} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( a+b \)

Explanation:
1. In a right triangle \( R=\frac c2 \) and \( r=\frac{a+b-c}{2} \).
2. So \( 2(r+R)=(a+b-c)+c=a+b \).

Teacher's Note:
1. Circumradius of a right triangle is half the hypotenuse.
2. Inradius of a right triangle: (a + b - c)/2.

Answer: (b) 1

Explanation:
1. \( \sin A\sin B\sin C+\cos A\cos B\le\sin A\sin B+\cos A\cos B=\cos(A-B)\le1 \).
2. Equality needs \( \sin C=1 \) (and \( A=B \)), so \( \sin C=1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( (\sqrt2+1):1 \)

Explanation:
1. Take legs 1 and hypotenuse \( \sqrt2 \).
2. Circumradius \( R=\frac{\sqrt2}{2} \) and inradius \( r=\frac{1+1-\sqrt2}{2} \).
3. Ratio \( \frac{R}{r}=\frac{\sqrt2}{2-\sqrt2}=\sqrt2+1 \).

Teacher's Note:
1. In a right triangle R is half the hypotenuse and r = (a + b - c)/2.
2. Rationalise the denominator.

Answer: (c) two

Explanation:
1. First: \( 2s^2-(1-2s^2)=0 \) with \( s=\sin\theta \) gives \( s=\pm\frac12 \).
2. Second: \( 2(1-s^2)-3s=0 \) gives \( 2s^2+3s-2=0 \), so \( s=\frac12 \) or \( -2 \) (rejected).
3. Common value \( \sin\theta=\frac12 \): \( \theta=\frac\pi6,\frac{5\pi}6 \) — two solutions.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( 2n\pi+\frac\pi4 \)

Explanation:
1. Rewrite as \( \log_{\sin x}\frac{\cos x}{\sin2x}=1 \), so \( \frac{1}{2\sin x}=\sin x \).
2. Then \( \sin^2x=\frac12 \) and \( \sin x=\frac1{\sqrt2} \) (the base must be positive and not 1).
3. Also \( \cos x>0 \), so \( x=2n\pi+\frac\pi4 \).

Teacher's Note:
1. The base of a logarithm must be positive and different from 1, and the argument positive.
2. These restrictions discard x = 3 pi/4.

Answer: (d) \( b+c=2a \) and the locus of \( A \) is an ellipse

Explanation:
1. \( \cos B+\cos C=2\sin\frac A2\cos\frac{B-C}2=4\sin^2\frac A2 \) gives \( \cos\frac{B-C}2=2\sin\frac A2 \).
2. Multiplying by \( 2\cos\frac A2 \): \( \sin B+\sin C=2\sin A \), i.e. \( b+c=2a \).
3. As \( BC=a \) is fixed, \( AB+AC=2a>a \) is constant: an ellipse.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \left(0,\frac\pi6\right)\cup\left(\frac{5\pi}6,2\pi\right) \)

Explanation:
1. \( 2\sin^2\theta-5\sin\theta+2=(2\sin\theta-1)(\sin\theta-2)>0 \).
2. Since \( \sin\theta-2<0 \), we need \( 2\sin\theta-1<0 \), i.e. \( \sin\theta<\frac12 \).
3. In \( (0,2\pi) \) this gives \( \theta\in\left(0,\frac\pi6\right)\cup\left(\frac{5\pi}6,2\pi\right) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 3

Explanation:
1. \( \frac12\cdot6\cdot10\sin C=15\sqrt3 \) gives \( \sin C=\frac{\sqrt3}2 \), so \( C=120^\circ \).
2. \( AB^2=36+100+60=196 \), so \( AB=14 \) and \( s=15 \).
3. Inradius \( r=\frac{\Delta}{s}=\sqrt3 \), so \( r^2=3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \frac\pi4 \) and \( \frac\pi2 \)

Explanation:
1. With \( c=\cos^2\theta \): \( 2(1-c)+4c(1-c)=2 \) gives \( 2c-4c^2=0 \), so \( c=0 \) or \( c=\frac12 \).
2. Hence \( \theta=\frac\pi2 \) or \( \theta=\frac\pi4,\frac{3\pi}4 \) (in \( (0,\pi) \); among the options, \( \frac\pi4 \) and \( \frac\pi2 \)).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( a,\ \sin A,\ R \)

Explanation:
1. With \( a,\sin A,R \): \( a=2R\sin A \), so the data is redundant and only one independent quantity (\( A \) and the size) is known; the shape is not fixed.
2. The other sets determine two angles or all three sides (acute angles are fixed by their sines), so the triangle is unique.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( -1\le\lambda\le-\frac34 \)

Explanation:
1. Let \( c=\cos^2\theta\in[0,1] \). Then \( -\lambda=c^2-c+1=\left(c-\tfrac12\right)^2+\tfrac34 \).
2. On \( [0,1] \) this ranges from \( \tfrac34 \) to \( 1 \).
3. So \( -1\le\lambda\le-\tfrac34 \).

Teacher's Note:
1. Convert everything to cos squared theta and find the range of the quadratic.
2. The vertex is at c = 1/2 and the maximum at the ends.

Answer: (a) \( \sin A\sin B\sin C \)

Explanation:
1. Factor \( \cos A\cos B\cos C \): the sum is \( \cos A\cos B\cos C\,(\tan A+\tan B+\tan C) \).
2. In a triangle \( \tan A+\tan B+\tan C=\tan A\tan B\tan C \).
3. So the sum equals \( \cos A\cos B\cos C\tan A\tan B\tan C=\sin A\sin B\sin C \).

Teacher's Note:
1. Use the triangle identity for the sum of tangents.
2. Divide and multiply by cos A cos B cos C.

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

Explanation:
1. Squaring: \( f^2=a^2+b^2+2\sqrt{P} \) with \( P=a^2b^2+(a^2-b^2)^2\sin^2x\cos^2x \).
2. So \( |ab|\le\sqrt P\le\frac{a^2+b^2}2 \).
3. Max \( f^2=2(a^2+b^2) \), min \( f^2=(|a|+|b|)^2 \); difference \( (|a|-|b|)^2=(a-b)^2 \) when \( ab\ge0 \).

Teacher's Note:
1. Expand the product under the root as a constant plus a multiple of sin squared x cos squared x.
2. sin^2 x cos^2 x ranges from 0 to 1/4.

Answer: (b) 3

Explanation:
1. \( \tan\theta=\cot5\theta \) gives \( \cos6\theta=0 \), so \( \theta=\frac\pi{12}+\frac{k\pi}6 \): \( \pm\frac\pi{12},\ \pm\frac{\pi}4,\ \pm\frac{5\pi}{12} \) with alternate signs as in the interval.
2. The second equation: \( 2\sin^22\theta+\sin2\theta-1=0 \), so \( \sin2\theta=\frac12 \) or \( -1 \), i.e. \( \theta=\frac\pi{12},\frac{5\pi}{12},-\frac\pi4 \).
3. All three satisfy both: 3 values.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 6

Explanation:
1. Each of sin and cos is at most 1, so both must equal 1.
2. In \( [0,4\pi] \): \( \sin\alpha=1 \) gives \( \alpha=\frac{\pi}{2},\frac{5\pi}{2} \) (2 values).
3. \( \cos\beta=1 \) gives \( \beta=0,2\pi,4\pi \) (3 values).
4. Pairs = 2 x 3 = 6.

Teacher's Note:
1. When a sum of bounded quantities reaches its maximum, each term must reach its own maximum.
2. Count the solutions of each equation separately and multiply.

Answer: (d) circumcentre

Explanation:
1. Equal elevation angles from the corners mean the horizontal distances from the foot to each corner are equal, as \( \tan\theta=\frac{h}{d} \).
2. A point equidistant from the three vertices is the circumcentre.

Teacher's Note:
1. The height is common, so equal angles imply equal horizontal distances.
2. Equidistant from the vertices means circumcentre.