JEE Mathematics Trigonometric Functions MCQs Set 15

Answer: (b) \( \dfrac{63}{16} \)

Explanation:
1. As \( 0\lt\alpha+\beta\lt\dfrac\pi2 \) and \( 0\lt\alpha-\beta\lt\dfrac\pi4 \) (since \( \sin(\alpha-\beta)\gt0 \)), \( \tan(\alpha+\beta)=\dfrac43 \) and \( \tan(\alpha-\beta)=\dfrac5{12} \).
2. \( \tan2\alpha=\dfrac{\frac43+\frac5{12}}{1-\frac43\cdot\frac5{12}}=\dfrac{63}{16} \). Answer: (b)

Teacher's Note:
1. Write \( 2\alpha=(\alpha+\beta)+(\alpha-\beta) \).

Answer: (c) \( \dfrac1{16} \)

Explanation:
1. Use \( \sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\dfrac14\sin3\theta \) with \( \theta=10^\circ \): \( \sin10^\circ\sin50^\circ\sin70^\circ=\dfrac14\sin30^\circ=\dfrac18 \).
2. The product is \( \dfrac12\cdot\dfrac18=\dfrac1{16} \). Answer: (c)

Teacher's Note:
1. Group \( 10^\circ,50^\circ,70^\circ \) into the triple-angle product.

Answer: (d) third and fourth quadrants only

Explanation:
1. \( \sin x\sin(x+2)=\dfrac12[\cos2-\cos(2x+2)] \) and \( \sin^2(x+1)=\dfrac12[1-\cos(2x+2)] \).
2. So \( y=\dfrac12(\cos2-1)=-\sin^21 \), a negative constant: the line \( y=-\sin^21 \) lies below the \( x \)-axis, in the third and fourth quadrants only. Answer: (d)

Teacher's Note:
1. The \( x \)-dependence cancels, leaving a horizontal line.

Answer: (b) \( \dfrac74 \)

Explanation:
1. Triangle inequalities: \( 5+5r\gt5r^2 \) gives \( r^2-r-1\lt0 \), i.e. \( r\lt\dfrac{1+\sqrt5}2 \); \( 5r+5r^2\gt5 \) gives \( r^2+r-1\gt0 \), i.e. \( r\gt\dfrac{\sqrt5-1}2 \).
2. So \( 0.618\lt r\lt1.618 \). Of the options only \( \dfrac74=1.75 \) lies outside. Answer: (b)

Teacher's Note:
1. Apply the triangle inequality to each pair of sides.

Answer: (a) 7:1

Explanation:
1. \( C=60^\circ \). Napier's rule: \( \tan\dfrac{A-B}2=\dfrac{a-b}{a+b}\cot\dfrac C2=\dfrac2{2\sqrt3}\cdot\sqrt3=1 \).
2. So \( A-B=90^\circ \); with \( A+B=120^\circ \), \( A=105^\circ \), \( B=15^\circ \), ratio \( 7:1 \). Answer: (a)

Teacher's Note:
1. Use the tangent rule (Napier's analogy).

Answer: (d) 25

Explanation:
1. Area \( =\dfrac{3\sqrt3}4r^2=27\sqrt3 \) gives \( r^2=36 \), \( r=6 \) (circumradius).
2. For the circle, \( r^2=25+36-c=36 \), so \( c=25 \). Answer: (d)

Teacher's Note:
1. For an equilateral triangle, area \( =\dfrac{3\sqrt3}4R^2 \).

Answer: (b) \( \dfrac c{\sqrt3} \)

Explanation:
1. \( (a+b)^2-c^2=ab \) gives \( \cos C=\dfrac{a^2+b^2-c^2}{2ab}=-\dfrac12 \), so \( C=120^\circ \).
2. \( R=\dfrac c{2\sin C}=\dfrac c{\sqrt3} \). Answer: (b)

Teacher's Note:
1. Use the cosine rule, then the sine rule.

Answer: (d) \( \sqrt{19} \)

Explanation:
1. \( 5\sin\left(\theta-\dfrac\pi6\right)=\dfrac{5\sqrt3}2\sin\theta-\dfrac52\cos\theta \), so the expression is \( \dfrac12\cos\theta+\dfrac{5\sqrt3}2\sin\theta \).
2. Maximum \( =\sqrt{\dfrac14+\dfrac{75}4}=\sqrt{19} \). Answer: (d)

Teacher's Note:
1. Use the maximum of \( a\cos\theta+b\sin\theta \), which is \( \sqrt{a^2+b^2} \).

Answer: (c) \( -\sqrt2 \)

Explanation:
1. AM-GM on \( \sin^4\alpha,4\cos^4\beta,1,1 \) gives \( \sin^4\alpha+4\cos^4\beta+2\ge4\sqrt2\sin\alpha\cos\beta \); equality forces \( \sin^4\alpha=4\cos^4\beta=1 \).
2. So \( \sin\alpha=1 \), \( \cos\beta=\dfrac1{\sqrt2} \) (positive, hence \( \sin\beta=\dfrac1{\sqrt2} \)). The value is \( -2\sin\alpha\sin\beta=-\sqrt2 \). Answer: (c)

Teacher's Note:
1. Equality in AM-GM fixes the values.

Answer: (a) \( \dfrac{(p^2+q^2)\sin\theta}{p\cos\theta+q\sin\theta} \)

Explanation:
1. \( BD=\sqrt{p^2+q^2} \). In \( \triangle ABD \), \( \angle ABD=\alpha \) where \( \alpha=\angle BDC \) (alternate angles, \( AB\parallel CD \)), so \( \sin\alpha=\dfrac p{\sqrt{p^2+q^2}} \), \( \cos\alpha=\dfrac q{\sqrt{p^2+q^2}} \); and \( \angle A=\pi-(\theta+\alpha) \).
2. Sine rule: \( AB=\dfrac{BD\sin\theta}{\sin(\theta+\alpha)}=\dfrac{(p^2+q^2)\sin\theta}{q\sin\theta+p\cos\theta} \). Answer: (a)

Teacher's Note:
1. Apply the sine rule in \( \triangle ABD \).

Answer: (b) \( \sec A\,\mathrm{cosec}A+1 \)

Explanation:
1. \( \dfrac{\tan A}{1-\cot A}=\dfrac{\sin^2A}{\cos A(\sin A-\cos A)\cdot\frac{1}{1}}\cdot\dfrac1{\,1\,}\cdot\dfrac{\cos A}{\cos A} \) simplifies to \( \dfrac{\sin^2A}{\cos A(\sin A-\cos A)} \), and similarly the second term is \( -\dfrac{\cos^2A}{\sin A(\sin A-\cos A)} \).
2. Sum \( =\dfrac{\sin^3A-\cos^3A}{\sin A\cos A(\sin A-\cos A)}=\dfrac{1+\sin A\cos A}{\sin A\cos A}=1+\sec A\,\mathrm{cosec}A \). Answer: (b)

Teacher's Note:
1. Convert to sines and cosines.

Answer: (d) \( \dfrac34\le A\le1 \)

Explanation:
1. \( A=1-\cos^2x+\cos^4x=\left(\cos^2x-\dfrac12\right)^2+\dfrac34 \).
2. As \( \cos^2x\in[0,1] \), \( \left(\cos^2x-\dfrac12\right)^2\in\left[0,\dfrac14\right] \), so \( \dfrac34\le A\le1 \). Answer: (d)

Teacher's Note:
1. Complete the square in \( \cos^2x \).

Answer: (b) \( \dfrac{56}{33} \)

Explanation:
1. Both angles are in the first quadrant: \( \tan(\alpha+\beta)=\dfrac34 \), \( \tan(\alpha-\beta)=\dfrac5{12} \).
2. \( \tan2\alpha=\dfrac{\frac34+\frac5{12}}{1-\frac34\cdot\frac5{12}}=\dfrac{14/12}{33/48}=\dfrac{56}{33} \). Answer: (b)

Teacher's Note:
1. \( 2\alpha=(\alpha+\beta)+(\alpha-\beta) \).

Answer: (c) Both \( A \) and \( B \) are true

Explanation:
1. Doubling the given relation and adding \( 3=\sum(\sin^2+\cos^2) \) gives \( (\sin\alpha+\sin\beta+\sin\gamma)^2+(\cos\alpha+\cos\beta+\cos\gamma)^2=0 \).
2. A sum of two squares is zero only if both are zero, so both \( A \) and \( B \) hold. Answer: (c)

Teacher's Note:
1. Expand the squares of the two sums.

Answer: (b) \( \dfrac12x^2 \)

Explanation:
1. Isosceles triangle with equal sides \( x \) and base angle \( \theta \): base \( =2x\cos\theta \), height \( =x\sin\theta \).
2. Area \( =\dfrac12x^2\sin2\theta \), maximum \( \dfrac12x^2 \). Answer: (b)

Teacher's Note:
1. Express the area in the base angle.

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

Explanation:
1. For a right angle at \( C \): \( c=2R \) and \( r=s-c=\dfrac{a+b-c}2 \), so \( 2r=a+b-c \).
2. Adding: \( 2r+2R=a+b-c+c=a+b \). Answer: (c)

Teacher's Note:
1. The hypotenuse is a diameter of the circumcircle.

Answer: (c) AP

Explanation:
1. Altitudes are \( \dfrac{2\Delta}a,\dfrac{2\Delta}b,\dfrac{2\Delta}c \); they are in HP exactly when \( \dfrac1a,\dots \) reciprocals i.e. \( a,b,c \) are in AP.
2. Since \( a:b:c=\sin A:\sin B:\sin C \), the sines are in AP. Answer: (c)

Teacher's Note:
1. Altitude \( =2\Delta/\text{side} \).

Answer: (c) \( 120^\circ \)

Explanation:
1. The largest side is \( c=\sqrt{1+\sin\alpha\cos\alpha} \) (\( c^2\gt1\gt\sin^2,\cos^2 \)).
2. \( \cos C=\dfrac{\sin^2\alpha+\cos^2\alpha-1-\sin\alpha\cos\alpha}{2\sin\alpha\cos\alpha}=-\dfrac12 \), so \( C=120^\circ \). Answer: (c)

Teacher's Note:
1. The greatest angle is opposite the greatest side.

Answer: (d) [-1, 3]

Explanation:
1. \( \sin x-\sqrt3\cos x=2\sin\left(x-\dfrac\pi3\right)\in[-2,2] \).
2. So \( f(x)\in[-1,3] \), and for \( f \) to be onto, \( S=[-1,3] \). Answer: (d)

Teacher's Note:
1. The codomain must equal the range.

Answer: (a) are in AP

Explanation:
1. With \( \cos^2\dfrac C2=\dfrac{s(s-c)}{ab} \) and \( \cos^2\dfrac A2=\dfrac{s(s-a)}{bc} \): \( \dfrac{s(s-c)+s(s-a)}b=\dfrac{3b}2 \).
2. \( s(2s-a-c)=\dfrac{3b^2}2 \) gives \( s\cdot b=\dfrac{3b^2}2 \), so \( a+b+c=3b \), i.e. \( 2b=a+c \). Answer: (a)

Teacher's Note:
1. Use the half-angle formulas in terms of \( s \).

Answer: (c) \( \dfrac{32}{3\sqrt3} \) sq units

Explanation:
1. The centroid \( G \) gives \( AG=\dfrac83 \). In \( \triangle ABG \), \( \angle A=\dfrac\pi6 \), \( \angle B=\dfrac\pi3 \), so \( \angle G=\dfrac\pi2 \) and \( BG=AG\cot\dfrac\pi3=\dfrac8{3\sqrt3} \).
2. \( [ABD]=\dfrac12\cdot AD\cdot BG=\dfrac{16}{3\sqrt3} \) and the median halves the area, so \( [ABC]=\dfrac{32}{3\sqrt3} \). Answer: (c)

Teacher's Note:
1. The centroid divides the median in the ratio \( 2:1 \).

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

Explanation:
1. \( \dfrac{A-B+C}2=\dfrac\pi2-B \), so the expression is \( 2ca\cos B \).
2. \( 2ca\cos B=c^2+a^2-b^2 \) by the cosine rule. Answer: (b)

Teacher's Note:
1. Use \( A+B+C=\pi \).

Answer: (c) \( \dfrac{\sqrt3}2 \)

Explanation:
1. \( \dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\cos2\theta \).
2. With \( \theta=15^\circ \): \( \cos30^\circ=\dfrac{\sqrt3}2 \). Answer: (c)

Teacher's Note:
1. Recognise the double-angle formula for cosine.

Answer: (b) \( -\dfrac45 \) or \( \dfrac45 \)

Explanation:
1. \( \tan\theta\lt0 \) puts \( \theta \) in the second or fourth quadrant.
2. In the second quadrant \( \sin\theta=\dfrac45 \); in the fourth \( \sin\theta=-\dfrac45 \). Both occur. Answer: (b)

Teacher's Note:
1. The sign of tangent allows two quadrants.

Answer: (a) 1

Explanation:
1. \( \alpha+\beta=\dfrac\pi2 \) (principal), so \( \tan(\alpha+2\beta)\tan(2\alpha+\beta)=\tan\left(\dfrac\pi2+\beta\right)\tan\left(\dfrac\pi2+\alpha\right)=\cot\alpha\cot\beta \).
2. With \( \alpha=\dfrac\pi3 \), \( \beta=\dfrac\pi6 \): \( \cot\dfrac\pi3\cot\dfrac\pi6=\dfrac1{\sqrt3}\cdot\sqrt3=1 \). Answer: (a)

Teacher's Note:
1. Solve for \( \alpha \) and \( \beta \) first.

Answer: (d) \( y\ge2 \)

Explanation:
1. \( y=(\sin\theta-\mathrm{cosec}\theta)^2+2 \).
2. Hence \( y\ge2 \) (AM-GM also gives this). Answer: (d)

Teacher's Note:
1. Complete the square.

Answer: (a) \( c^2-3c-7=0 \)

Explanation:
1. \( \cos60^\circ=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{c^2-7}{6c} \).
2. \( \dfrac12=\dfrac{c^2-7}{6c} \) gives \( c^2-3c-7=0 \). Answer: (a)

Teacher's Note:
1. Apply the cosine rule.

Answer: (b) \( a,b \) and \( c \) are in AP

Explanation:
1. \( \tan\dfrac A2\tan\dfrac C2=\dfrac13 \) and also equals \( \dfrac{s-b}s \), so \( \dfrac{s-b}s=\dfrac13 \).
2. \( 2s=3b \), i.e. \( a+b+c=3b \), so \( a+c=2b \): \( a,b,c \) are in AP. Answer: (b)

Teacher's Note:
1. Use \( \tan\frac A2\tan\frac C2=\frac{s-b}s \).