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NCERT Exemplar · Class 11 Mathematics Trigonometric Functions

76 questions · 76 still being checked

EXERCISE 3.3 71–76 (part 8 of 8)

  1. In each of the Exercises $\displaystyle 68$ to $\displaystyle 75$, state whether the statements is True or False? Also give justification.

    Exercise 71

    cos⁡2π15cos⁡4π15cos⁡8π15cos⁡16π15=116\displaystyle \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}

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    NCERT’s answer
    True
    TrueLet \(\displaystyle \theta=\frac{2\pi}{15}\); the angles are \(\displaystyle \theta,2\theta,4\theta,8\theta\). Using \(\displaystyle \sin2\alpha=2\sin\alpha\cos\alpha\) repeatedly: \[16\sin\theta\cos\theta\cos2\theta\cos4\theta\cos8\theta=\sin16\theta \] \[16\theta=\tfrac{32\pi}{15}=2\pi+\tfrac{2\pi}{15} \;\Rightarrow\; \sin16\theta=\sin\theta \] \[\cos\theta\cos2\theta\cos4\theta\cos8\theta=\frac{\sin\theta}{16\sin\theta}=\frac1{16} \]

  2. Exercise 72

    One value of θ\displaystyle \theta which satisfies the equation sin⁡4θ−2sin⁡2θ−1\displaystyle \sin ^4 \theta-2 \sin ^2 \theta-1 lies between 0\displaystyle 0 and 2π\displaystyle 2 \pi.

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    NCERT’s answer
    False
    FalseSet \(\displaystyle \sin^4\theta-2\sin^2\theta-1=0\) and put \(\displaystyle t=\sin^2\theta\). \[t^2-2t-1=0 \;\Rightarrow\; t=1\pm\sqrt2 \] \[1+\sqrt2>1,\qquad 1-\sqrt2<0 \] \[0\le\sin^2\theta\le1 \] Neither root is possible, so no \(\displaystyle \theta\) satisfies it.

  3. Exercise 73

    If cosec⁡x=1+cot⁡x\displaystyle \operatorname{cosec} x=1+\cot x then x=2nπ,2nπ+π2\displaystyle x=2 n \pi, 2 n \pi+\frac{\pi}{2}

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    NCERT’s answer
    True
    True. Every solution has one of these forms, but \(\displaystyle x = 2n\pi\) is not itself a solution; the solutions are exactly \(\displaystyle x = 2n\pi + \frac{\pi}{2}\).\[\frac{1}{\sin x} = 1 + \frac{\cos x}{\sin x} \implies \sin x + \cos x = 1 \quad (\sin x \ne 0) \] \[\sqrt2 \sin\left(x + \frac{\pi}{4}\right) = 1 \implies x + \frac{\pi}{4} = 2n\pi + \frac{\pi}{4} \ \text{or}\ 2n\pi + \frac{3\pi}{4} \] \[x = 2n\pi \quad \text{or} \quad x = 2n\pi + \frac{\pi}{2} \] \[x = 2n\pi:\ \sin x = 0 \implies \operatorname{cosec} x,\ \cot x \ \text{undefined} \] \[x = 2n\pi + \frac{\pi}{2}:\ \operatorname{cosec} x = 1 = 1 + 0 = 1 + \cot x \]

  4. Exercise 74

    If tan⁡θ+tan⁡2θ+3tan⁡θtan⁡2θ=3\displaystyle \tan \theta+\tan 2 \theta+\sqrt{3} \tan \theta \tan 2 \theta=\sqrt{3}, then θ=nπ3+π9\displaystyle \theta=\frac{n \pi}{3}+\frac{\pi}{9}

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    NCERT’s answer
    True
    True.\[\tan\theta + \tan 2\theta = \sqrt3\,(1 - \tan\theta\tan 2\theta) \] \[1 - \tan\theta\tan2\theta = 0 \implies \tan 2\theta = -\tan\theta \implies \tan^2\theta = -1 \quad \text{(impossible)} \] \[\tan 3\theta = \frac{\tan\theta + \tan 2\theta}{1 - \tan\theta\tan 2\theta} = \sqrt3 = \tan\frac{\pi}{3} \] \[3\theta = n\pi + \frac{\pi}{3} \] \[\theta = \frac{n\pi}{3} + \frac{\pi}{9} \]Answer: True.

  5. Exercise 75

    If tan⁡(πcos⁡θ)=cot⁡(πsin⁡θ)\displaystyle \tan (\pi \cos \theta)=\cot (\pi \sin \theta), then cos⁡(θ−π4)=±122\displaystyle \cos \left(\theta-\frac{\pi}{4}\right)= \pm \frac{1}{2 \sqrt{2}}

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    NCERT’s answer
    True
    True. The condition forces \(\displaystyle \sin\theta + \cos\theta = \pm\frac12\).\[\tan(\pi\cos\theta) = \cot(\pi\sin\theta) = \tan\left(\frac{\pi}{2} - \pi\sin\theta\right) \] \[\pi\cos\theta = n\pi + \frac{\pi}{2} - \pi\sin\theta \implies \sin\theta + \cos\theta = \frac{2n+1}{2} \] \[|\sin\theta + \cos\theta| \le \sqrt2 \implies |2n+1| \le 2\sqrt2 < 3 \implies 2n+1 = \pm 1 \] \[\cos\left(\theta - \frac{\pi}{4}\right) = \frac{\cos\theta + \sin\theta}{\sqrt2} = \pm\frac{1}{2\sqrt2} \]Answer: True.

  6. Exercise 76

    In the following match each item given under the column C1\displaystyle \mathrm{C}_1 to its correct answer given under the column C2\displaystyle \mathrm{C}_2 :
    (a) sin⁡(x+y)sin⁡(x−y)\displaystyle \sin (x+y) \sin (x-y)(i) cos⁡2x−sin⁡2y\displaystyle \cos ^2 x-\sin ^2 y
    (b) cos⁡(x+y)cos⁡(x−y)\displaystyle \cos (x+y) \cos (x-y)(ii) 1−tan⁡θ1+tan⁡θ\displaystyle \frac{1-\tan \theta}{1+\tan \theta}
    (c) cot⁡(π4+θ)\displaystyle \cot \left(\frac{\pi}{4}+\theta\right)(iii) 1+tan⁡θ1−tan⁡θ\displaystyle \frac{1+\tan \theta}{1-\tan \theta}
    (d) tan⁡(π4+θ)\displaystyle \tan \left(\frac{\pi}{4}+\theta\right)(iv) sin⁡2x−sin⁡2y\displaystyle \sin ^2 x-\sin ^2 y

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    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (iv)
    (b)
    \(\displaystyle \leftrightarrow\) (i)
    (c)
    \(\displaystyle \leftrightarrow\) (ii)
    (d)
    \(\displaystyle \leftrightarrow\) (iii)
    (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
    (a)
    \[\sin(x+y)\sin(x-y) = \sin^2 x - \sin^2 y \]
    (b)
    \[\cos(x+y)\cos(x-y) = \cos^2 x - \sin^2 y \]
    (c)
    \[\cot\left(\frac{\pi}{4}+\theta\right) = \frac{1 - \tan\theta}{1 + \tan\theta} \]
    (d)
    \[\tan\left(\frac{\pi}{4}+\theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \]