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NCERT Exemplar · Class 11 Mathematics Trigonometric Functions

76 questions · 76 still being checked

EXERCISE 3.3 41–50 (part 5 of 8)

  1. Choose the correct answer from the given four options in the Exercises $\displaystyle 30$ to $\displaystyle 59$ (M.C.Q.).

    Exercise 41

    The minimum value of 3cos⁡x+4sin⁡x+8\displaystyle 3 \cos x+4 \sin x+8 is
    (A)
    5\displaystyle 5 (B) 9\displaystyle 9 (C) 7\displaystyle 7 (D) 3\displaystyle 3

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    NCERT’s answer
    D
    (D) \(\displaystyle 3\)The amplitude of \(\displaystyle 3\cos x + 4\sin x\) is \(\displaystyle \sqrt{3^2+4^2}\).\[3\cos x + 4\sin x = 5\cos(x-\alpha), \quad \cos\alpha = \tfrac35,\ \sin\alpha = \tfrac45 \]\[-5 \le 3\cos x + 4\sin x \le 5 \]\[\text{minimum} = 8 - 5 = 3 \]

  2. Exercise 42

    The value of tan⁡3 A−tan⁡2 A−tan⁡A\displaystyle \tan 3 \mathrm{~A}-\tan 2 \mathrm{~A}-\tan \mathrm{A} is equal to
    (A)
    tan⁡3 Atan⁡2 Atan⁡ A\displaystyle \tan 3 \mathrm{~A} \tan 2 \mathrm{~A} \tan \mathrm{~A}
    (B)
    −tan⁡3 Atan⁡2 Atan⁡ A\displaystyle -\tan 3 \mathrm{~A} \tan 2 \mathrm{~A} \tan \mathrm{~A}
    (C)
    tan⁡Atan⁡2 A−tan⁡2 Atan⁡3 A−tan⁡3 Atan⁡A\displaystyle \tan \mathrm{A} \tan 2 \mathrm{~A}-\tan 2 \mathrm{~A} \tan 3 \mathrm{~A}-\tan 3 \mathrm{~A} \tan \mathrm{A}
    (D)
    None of these

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    NCERT’s answer
    A
    (A) \(\displaystyle \tan 3A\tan 2A\tan A\)\[\tan 3A = \tan(2A + A) = \frac{\tan 2A + \tan A}{1 - \tan 2A\tan A} \]\[\tan 3A - \tan 3A\tan 2A\tan A = \tan 2A + \tan A \]\[\tan 3A - \tan 2A - \tan A = \tan 3A\tan 2A\tan A \]

  3. Exercise 43

    The value of sin⁡(45∘+θ)−cos⁡(45∘−θ)\displaystyle \sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right) is
    (A)
    2cos⁡θ\displaystyle 2 \cos \theta
    (B)
    2sin⁡θ\displaystyle 2 \sin \theta
    (C)
    1\displaystyle 1 (D) 0\displaystyle 0

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    NCERT’s answer
    D
    (D) \(\displaystyle 0\)\[\sin(45^\circ + \theta) = \cos\bigl(90^\circ - (45^\circ + \theta)\bigr) = \cos(45^\circ - \theta) \]\[\sin(45^\circ + \theta) - \cos(45^\circ - \theta) = 0 \]

  4. Exercise 44

    The value of cot⁡(π4+θ)cot⁡(π4−θ)\displaystyle \cot \left(\frac{\pi}{4}+\theta\right) \cot \left(\frac{\pi}{4}-\theta\right) is
    (A)
    -1\displaystyle 1 (B) 0\displaystyle 0 (C) 1\displaystyle 1 (D) Not defined

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    NCERT’s answer
    C
    (C) \(\displaystyle 1\)\[\cot\left(\frac{\pi}{4} - \theta\right) = \tan\left(\frac{\pi}{2} - \frac{\pi}{4} + \theta\right) = \tan\left(\frac{\pi}{4} + \theta\right) \]\[\cot\left(\frac{\pi}{4} + \theta\right)\tan\left(\frac{\pi}{4} + \theta\right) = 1 \]

  5. Exercise 45

    cos⁡2θcos⁡2ϕ+sin⁡2(θ−ϕ)−sin⁡2(θ+ϕ)\displaystyle \cos 2 \theta \cos 2 \phi+\sin ^2(\theta-\phi)-\sin ^2(\theta+\phi) is equal to
    (A)
    sin⁡2(θ+ϕ)\displaystyle \sin 2(\theta+\phi)
    (B)
    cos⁡2(θ+ϕ)\displaystyle \cos 2(\theta+\phi)
    (C)
    sin⁡2(θ−ϕ)\displaystyle \sin 2(\theta-\phi)
    (D)
    cos⁡2(θ−ϕ)\displaystyle \cos 2(\theta-\phi)
    [Hint: Use sin⁡2 A−sin⁡2 B=sin⁡(A+B)sin⁡(A−B)\displaystyle \sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}=\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B}) ]

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    NCERT’s answer
    B
    (B) \(\displaystyle \cos 2(\theta+\phi)\)Take \(\displaystyle A = \theta - \phi,\ B = \theta + \phi\) in the hint.\[\sin^2(\theta-\phi) - \sin^2(\theta+\phi) = \sin 2\theta\,\sin(-2\phi) = -\sin 2\theta\sin 2\phi \]\[\text{Expression} = \cos 2\theta\cos 2\phi - \sin 2\theta\sin 2\phi = \cos(2\theta + 2\phi) \]

  6. Exercise 46

    The value of cos⁡12∘+cos⁡84∘+cos⁡156∘+cos⁡132∘\displaystyle \cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ} is
    (A)
    12\displaystyle \frac{1}{2}
    (B)
    1\displaystyle 1 (C) −12\displaystyle -\frac{1}{2}
    (D)
    18\displaystyle \frac{1}{8}

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    NCERT’s answer
    C
    (C) \(\displaystyle -\tfrac12\)\[\cos 12^\circ + \cos 132^\circ = 2\cos 72^\circ\cos 60^\circ = \cos 72^\circ \]\[\cos 84^\circ + \cos 156^\circ = 2\cos 120^\circ\cos 36^\circ = -\cos 36^\circ \]\[\cos 72^\circ = \sin 18^\circ = \frac{\sqrt5 - 1}{4}, \qquad \cos 36^\circ = \frac{\sqrt5 + 1}{4} \]\[\text{Sum} = \cos 72^\circ - \cos 36^\circ = \frac{\sqrt5 - 1}{4} - \frac{\sqrt5 + 1}{4} = -\frac12 \]

  7. Exercise 47

    If tan⁡A=12,tan⁡ B=13\displaystyle \tan \mathrm{A}=\frac{1}{2}, \tan \mathrm{~B}=\frac{1}{3}, then tan⁡(2 A+B)\displaystyle \tan (2 \mathrm{~A}+\mathrm{B}) is equal to
    (A)
    1\displaystyle 1 (B) 2\displaystyle 2 (C) 3\displaystyle 3 (D) 4\displaystyle 4

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    NCERT’s answer
    C
    (C) \(\displaystyle 3\)\[\tan 2A = \frac{2\tan A}{1 - \tan^2 A} = \frac{1}{1 - \frac14} = \frac43 \]\[\tan(2A + B) = \frac{\tan 2A + \tan B}{1 - \tan 2A\tan B} = \frac{\frac43 + \frac13}{1 - \frac49} = \frac{5/3}{5/9} = 3 \]

  8. Exercise 48

    The value of sin⁡π10sin⁡13π10\displaystyle \sin \frac{\pi}{10} \sin \frac{13 \pi}{10} is
    (A)
    12\displaystyle \frac{1}{2}
    (B)
    −12\displaystyle -\frac{1}{2}
    (C)
    −14\displaystyle -\frac{1}{4}
    (D)
    1\displaystyle 1 [Hint: Use sin⁡18∘=5−14\displaystyle \sin 18^{\circ}=\frac{\sqrt{5}-1}{4} and cos⁡36∘=5+14\displaystyle \cos 36^{\circ}=\frac{\sqrt{5}+1}{4} ]

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    NCERT’s answer
    C
    (C) \(\displaystyle -\tfrac14\)\[\frac{13\pi}{10} = 234^\circ, \qquad \sin 234^\circ = \sin(180^\circ + 54^\circ) = -\sin 54^\circ = -\cos 36^\circ \]\[\sin 18^\circ\sin 234^\circ = -\sin 18^\circ\cos 36^\circ = -\frac{\sqrt5 - 1}{4}\cdot\frac{\sqrt5 + 1}{4} \]\[= -\frac{5 - 1}{16} = -\frac14 \]

  9. Exercise 49

    The value of sin⁡50∘−sin⁡70∘+sin⁡10∘\displaystyle \sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ} is equal to
    (A)
    1\displaystyle 1 (B) 0\displaystyle 0 (C) 12\displaystyle \frac{1}{2}

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    NCERT’s answer
    B
    (B) \(\displaystyle 0\)\[\sin 50^\circ - \sin 70^\circ = 2\cos 60^\circ \sin(-10^\circ) = -\sin 10^\circ \]\[\sin 50^\circ - \sin 70^\circ + \sin 10^\circ = -\sin 10^\circ + \sin 10^\circ = 0 \]

  10. Exercise 50

    If sin⁡θ+cos⁡θ=1\displaystyle \sin \theta+\cos \theta=1, then the value of sin⁡2θ\displaystyle \sin 2 \theta is equal to
    (A)
    1\displaystyle 1 (B) 12\displaystyle \frac{1}{2}
    (C)
    0\displaystyle 0 (D) -1\displaystyle 1

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    NCERT’s answer
    C
    (C) \(\displaystyle 0\)Square both sides.\[(\sin\theta + \cos\theta)^2 = 1 \]\[1 + \sin 2\theta = 1 \]\[\sin 2\theta = 0 \]