JEE Mathematics Trigonometrical Functions and Identities MCQs Set 07

Answer: (b) 23

Explanation:
1. If \(A+B=45^\circ\), then \((1+\tan A)(1+\tan B)=2\).
2. Pairs \((1,44),(2,43),\dots,(22,23)\) give \(2^{22}\), and \(1+\tan45^\circ=2\). Total \(2^{23}\), so \(n=23\).

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Answer: (d) \(\dfrac12\operatorname{cosec}\alpha\left[\tan(n+1)\alpha-\tan\alpha\right]\)

Explanation:
1. \(\cos\alpha+\cos(2r+1)\alpha=2\cos(r+1)\alpha\cos r\alpha\), and \(\dfrac{\sin\alpha}{\cos r\alpha\cos(r+1)\alpha}=\tan(r+1)\alpha-\tan r\alpha\).
2. So each term is \(\dfrac1{2\sin\alpha}\left[\tan(r+1)\alpha-\tan r\alpha\right]\) and the sum telescopes to \(\dfrac12\operatorname{cosec}\alpha\left[\tan(n+1)\alpha-\tan\alpha\right]\).

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Answer: (a) [0, 2]

Explanation:
1. \(\sin x-\cos x=\sqrt2\sin\left(x-\dfrac\pi4\right)\), so \(\sqrt2(\sin x-\cos x)\in[-2,2]\) and the bracket lies in \([1,5]\).
2. \(\log_{\sqrt5}1=0\) and \(\log_{\sqrt5}5=2\), so the range is \([0,2]\).

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Answer: (b) \(\dfrac{2\cos x}{\sqrt{\cos2x}}\)

Explanation:
1. The sum is \(\dfrac{(a+b)+(a-b)}{\sqrt{a^2-b^2}}=\dfrac{2a}{\sqrt{a^2-b^2}}=\dfrac2{\sqrt{1-\tan^2x}}\).
2. \(\dfrac2{\sqrt{1-\tan^2x}}=\dfrac{2\cos x}{\sqrt{\cos^2x-\sin^2x}}=\dfrac{2\cos x}{\sqrt{\cos2x}}\).

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Answer: (a) (1) and (4) only

Explanation:
1. \(\sec\theta+\tan\theta=1\) and \(\sec\theta-\tan\theta=\dfrac1{\sec\theta+\tan\theta}=1\), so \(\sec\theta=1\), \(\tan\theta=0\), \(\cos\theta=1\), \(\sin\theta=0\).
2. The coefficients add to \(0\), so \(x=1\) is a root. Hence \(\sec\theta\) and \(\cos\theta\) are roots, while \(\tan\theta=\sin\theta=0\) are not (in general).

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Answer: (c) -2

Explanation:
1. \(\cos2001\pi=-1\); \(\cot\dfrac{2001\pi}2=\cot\left(1000\pi+\dfrac\pi2\right)=0\); \(\sec667\pi=-1\); \(\tan\dfrac{2001\pi}4=\tan\left(500\pi+\dfrac\pi4\right)=1\); \(\operatorname{cosec}\dfrac{2001\pi}6=\operatorname{cosec}\left(333\pi+\dfrac\pi2\right)=-1\).
2. Sum \(=-1+0-1+1-1=-2\).

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Answer: (a) \(\cos\alpha\)

Explanation:
1. Let \(T=\tan\dfrac\beta2\), so \(\tan\dfrac\alpha2=\dfrac T2\) and \(T^2=\dfrac{1-\cos\beta}{1+\cos\beta}\).
2. \(\cos\alpha=\dfrac{1-T^2/4}{1+T^2/4}=\dfrac{4-T^2}{4+T^2}=\dfrac{4(1+\cos\beta)-(1-\cos\beta)}{4(1+\cos\beta)+(1-\cos\beta)}=\dfrac{3+5\cos\beta}{5+3\cos\beta}\).

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Answer: (a) (1) and (4) only

Explanation:
1. \(a^2+b^2=2+2\cos(\theta-\phi)=4\cos^2\dfrac{\theta-\phi}2\).
2. So \(\cos\dfrac{\theta-\phi}2=\pm\dfrac12\sqrt{a^2+b^2}\) (statement 1) and \(\cos(\theta-\phi)=\dfrac{a^2+b^2-2}2\) (statement 4). Statements 2 and 3 do not follow.

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

Explanation:
1. Write \(\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\), so \(a-b\cos2\theta=\dfrac{(a-b)+(a+b)\tan^2\theta}{1+\tan^2\theta}\). Do the same for \(\varphi\).
2. With \(\tan^2\theta\tan^2\varphi=\dfrac{a-b}{a+b}\), the product simplifies to \(a^2-b^2\) (confirmed numerically for sample values).

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Answer: (d) 9

Explanation:
1. The expression equals \(\sin^2+\cos^2+2+2+\operatorname{cosec}^2+\sec^2=5+\sec^2\theta+\operatorname{cosec}^2\theta=7+\tan^2\theta+\cot^2\theta\).
2. \(\tan^2\theta+\cot^2\theta\ge2\), so the minimum is \(9\).

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Answer: (b) \(\dfrac1{2^n}\cot\dfrac\alpha{2^n}\)

Explanation:
1. Use \(\tan\theta=\cot\theta-2\cot2\theta\): \(\dfrac1{2^k}\tan\dfrac\alpha{2^k}=\dfrac1{2^k}\cot\dfrac\alpha{2^k}-\dfrac1{2^{k-1}}\cot\dfrac\alpha{2^{k-1}}\).
2. The sum from \(k=1\) to \(n\) telescopes to \(\dfrac1{2^n}\cot\dfrac\alpha{2^n}-\cot\alpha\). Adding \(\cot\alpha\) gives \(\dfrac1{2^n}\cot\dfrac\alpha{2^n}\).

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Answer: (b) Both A and B are true

Explanation:
1. \((\sum\cos)^2+(\sum\sin)^2=3+2\left[\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)\right]=3-3=0\).
2. A sum of two squares is zero only if each is zero, so both A and B are true.

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Answer: (b) \(\left]\dfrac\pi4,\dfrac\pi2\right[\)

Explanation:
1. \(f(x)=1-\dfrac12\sin^22x\), so \(f'(x)=-\sin4x\).
2. \(f'>0\) when \(\sin4x \lt 0\), i.e. \(4x\in(\pi,2\pi)\), \(x\in\left(\dfrac\pi4,\dfrac\pi2\right)\).

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Answer: (b) 54

Explanation:
1. Let the tower be \(OT\) of height \(h\). \(OA=h\cot45^\circ=h\) and \(OB=h\cot30^\circ=h\sqrt3\).
2. \(OA\) points west from \(A\) and \(AB\) points south, so the angle at \(A\) is \(90^\circ\): \(AB^2=OB^2-OA^2=2h^2\).
3. \(h\sqrt2=54\sqrt2\Rightarrow h=54\) m.

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Answer: (b) \((105^\circ,15^\circ)\)

Explanation:
1. By the tangent rule: \(\tan\dfrac{A-B}2=\dfrac{a-b}{a+b}\cot\dfrac C2=\dfrac{1+\sqrt3}{3+\sqrt3}\cdot\sqrt3=1\).
2. So \(A-B=90^\circ\); with \(A+B=120^\circ\), \(A=105^\circ\), \(B=15^\circ\).

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Answer: (b) \(\dfrac{h\cos\alpha-a\sin\alpha}{9\sin\alpha}\)

Explanation:
1. If the common spacing is \(d\), the longest pole is at distance \(a+9d\) from \(O\), and its top subtends \(\alpha\): \(h=(a+9d)\tan\alpha\).
2. \(d=\dfrac{h\cot\alpha-a}9=\dfrac{h\cos\alpha-a\sin\alpha}{9\sin\alpha}\).

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Answer: (c) \(\dfrac32-2\sqrt3\)

Explanation:
1. \(f\left(-\dfrac{11\pi}6\right)=-f\left(\dfrac{11\pi}6\right)=-\left(3\sin\dfrac{11\pi}6+4\cos\dfrac{11\pi}6\right)\).
2. \(=-\left(3\cdot\left(-\tfrac12\right)+4\cdot\tfrac{\sqrt3}2\right)=\dfrac32-2\sqrt3\).

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Answer: (c) \(\dfrac\pi2\)

Explanation:
1. \(|\sin4x|\) has period \(\dfrac\pi4\) and \(|\cos2x|\) has period \(\dfrac\pi2\), so \(\dfrac\pi2\) is a period.
2. \(\dfrac\pi4\) is not: \(f\left(x+\dfrac\pi4\right)=|\sin4x|+|\sin2x|\ne f(x)\). The fundamental period is \(\dfrac\pi2\).

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Answer: (b) \(\dfrac94\)

Explanation:
1. With \(c=\cos2x\): \(\sin^22x=1-c^2\), \(\cos^4x=\dfrac{(1+c)^2}4\). The expression is \(4+\dfrac12(1-c^2)-\dfrac12(1+c)^2=4-c-c^2\).
2. On \(c\in[-1,1]\): maximum at \(c=-\tfrac12\) is \(\dfrac{17}4\); minimum at \(c=1\) is \(2\). \(M-m=\dfrac94\).

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Answer: (c) \(\dfrac75\)

Explanation:
1. \(2\cos\theta\cos\dfrac{\alpha-\beta}2=\dfrac32\) and \(2\sin\theta\cos\dfrac{\alpha-\beta}2=\dfrac12\), so \(\tan\theta=\dfrac13\).
2. \(\sin2\theta=\dfrac{2/3}{10/9}=\dfrac35\), \(\cos2\theta=\dfrac{8/9}{10/9}=\dfrac45\). Sum \(=\dfrac75\).

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