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NCERT Exemplar · Class 11 Mathematics Trigonometric Functions

76 questions · 76 still being checked

EXERCISE 3.3 31–40 (part 4 of 8)

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

    Exercise 31

    If f(x)=cos⁡2x+sec⁡2x\displaystyle f(x)=\cos ^2 x+\sec ^2 x, then
    (A)
    f(x)<1\displaystyle f(x)<1
    (B)
    f(x)=1\displaystyle f(x)=1
    (C)
    2<f(x)<1\displaystyle 2<f(x)<1
    (D)
    f(x)≥2\displaystyle f(x) \geq 2
    [Hint: A.M≥G.M\displaystyle \mathrm{A} . \mathrm{M} \geq \mathrm{G} . \mathrm{M}.]

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    NCERT’s answer
    D
    (D) \(\displaystyle f(x)\ge 2\) \[\frac{\cos^2x+\sec^2x}{2} \ge \sqrt{\cos^2x\,\sec^2x} = 1 \quad \text{(A.M.} \ge \text{G.M.)} \] \[f(x) \ge 2 \]

  2. Exercise 32

    If tan⁡θ=12\displaystyle \tan \theta=\frac{1}{2} and tan⁡ϕ=13\displaystyle \tan \phi=\frac{1}{3}, then the value of θ+ϕ\displaystyle \theta+\phi is
    (A)
    π6\displaystyle \frac{\pi}{6}
    (B)
    π\displaystyle \pi
    (C)
    0\displaystyle 0 (D) π4\displaystyle \frac{\pi}{4}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{\pi}{4}\) \[\tan(\theta+\phi) = \frac{\tfrac12+\tfrac13}{1-\tfrac12\cdot\tfrac13} = \frac{5/6}{5/6} = 1 \] \[\theta+\phi = \tfrac{\pi}{4} \quad (\theta,\phi \text{ acute}) \]

  3. Exercise 33

    Which of the following is not correct?
    (A)
    sin⁡θ=−15\displaystyle \sin \theta=-\frac{1}{5}
    (B)
    cos⁡θ=1\displaystyle \cos \theta=1
    (C)
    sec⁡θ=12\displaystyle \sec \theta=\frac{1}{2}
    (D)
    tan⁡θ=20\displaystyle \tan \theta=20

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    NCERT’s answer
    C
    (C) \(\displaystyle \sec\theta=\frac{1}{2}\) is not possible. \[|\cos\theta|\le 1 \Rightarrow |\sec\theta|=\frac{1}{|\cos\theta|}\ge 1 \] \[\left|\tfrac12\right|<1 \] The other three values are attainable: \(\displaystyle |\sin\theta|\le 1\), \(\displaystyle \cos 0=1\), and \(\displaystyle \tan\theta\) takes every real value.

  4. Exercise 34

    The value of tan⁡1∘tan⁡2∘tan⁡3∘…tan⁡89∘\displaystyle \tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ} is
    (A)
    0\displaystyle 0 (B) 1\displaystyle 1 (C) 12\displaystyle \frac{1}{2}
    (D)
    Not defined

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    NCERT’s answer
    B
    (B) \(\displaystyle 1\). \[\tan(90^\circ-k^\circ)=\cot k^\circ \Rightarrow \tan k^\circ\,\tan(90^\circ-k^\circ)=1 \] Pair \(\displaystyle k\) with \(\displaystyle 90-k\) for \(\displaystyle k=1,\dots,44\): \[\text{Product}=1^{44}\cdot\tan 45^\circ=1 \]

  5. Exercise 35

    The value of 1−tan⁡215∘1+tan⁡215∘\displaystyle \frac{1-\tan ^2 15^{\circ}}{1+\tan ^2 15^{\circ}} is
    (A)
    1\displaystyle 1 (B) 3\displaystyle \sqrt{3}
    (C)
    32\displaystyle \frac{\sqrt{3}}{2}

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{\sqrt3}{2}\). \[\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}=\cos 2\theta \] \[\theta=15^\circ:\quad \cos 30^\circ=\frac{\sqrt3}{2} \]

  6. Exercise 36

    The value of cos⁡1∘cos⁡2∘cos⁡3∘…cos⁡179∘\displaystyle \cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots \cos 179^{\circ} is
    (A)
    12\displaystyle \frac{1}{\sqrt{2}}
    (B)
    0\displaystyle 0 (C) 1\displaystyle 1 (D) -1\displaystyle 1

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    NCERT’s answer
    B
    (B) \(\displaystyle 0\). The product contains the factor \(\displaystyle \cos 90^\circ\): \[\cos 90^\circ=0 \] \[\cos1^\circ\cos2^\circ\cdots\cos 90^\circ\cdots\cos179^\circ=0 \]

  7. Exercise 37

    If tan⁡θ=3\displaystyle \tan \theta=3 and θ\displaystyle \theta lies in third quadrant, then the value of sin⁡θ\displaystyle \sin \theta is
    (A)
    110\displaystyle \frac{1}{\sqrt{10}}
    (B)
    −110\displaystyle -\frac{1}{\sqrt{10}}
    (C)
    −310\displaystyle \frac{-3}{\sqrt{10}}
    (D)
    310\displaystyle \frac{3}{\sqrt{10}}

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{-3}{\sqrt{10}}\). \[\sec^2\theta=1+\tan^2\theta=1+9=10 \] \[\cos\theta=-\frac{1}{\sqrt{10}} \quad \text{(third quadrant)} \] \[\sin\theta=\tan\theta\cos\theta=3\left(-\frac{1}{\sqrt{10}}\right)=-\frac{3}{\sqrt{10}} \]

  8. Exercise 38

    The value of tan⁡75∘−cot⁡75∘\displaystyle \tan 75^{\circ}-\cot 75^{\circ} is equal to
    (A)
    23\displaystyle 2 \sqrt{3}
    (B)
    2+3\displaystyle 2+\sqrt{3}
    (C)
    2−3\displaystyle 2-\sqrt{3}

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    NCERT’s answer
    A
    (A) \(\displaystyle 2\sqrt3\). \[\tan75^\circ=\tan(45^\circ+30^\circ)=\frac{1+\dfrac{1}{\sqrt3}}{1-\dfrac{1}{\sqrt3}}=\frac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3 \] \[\cot75^\circ=\frac{1}{2+\sqrt3}=2-\sqrt3 \] \[\tan75^\circ-\cot75^\circ=(2+\sqrt3)-(2-\sqrt3)=2\sqrt3 \]

  9. Exercise 39

    Which of the following is correct?
    (A)
    sin⁡1∘>sin⁡1\displaystyle \sin 1^{\circ}>\sin 1
    (B)
    sin⁡1∘<sin⁡1\displaystyle \sin 1^{\circ}<\sin 1
    (C)
    sin⁡1∘=sin⁡1\displaystyle \sin 1^{\circ}=\sin 1
    (D)
    sin⁡1∘=π18∘sin⁡1\displaystyle \sin 1^{\circ}=\frac{\pi}{18^{\circ}} \sin 1
    [Hint: 1\displaystyle 1 radian =180∘π=57∘30′\displaystyle =\frac{180^{\circ}}{\pi}=57^{\circ} 30^{\prime} approx]

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    NCERT’s answer
    B
    (B) \(\displaystyle \sin 1^\circ<\sin 1\), where \(\displaystyle 1\) is in radians. \[1^\circ=\frac{\pi}{180}\ \text{rad}\approx 0.0175 \] \[0<\frac{\pi}{180}<1<\frac{\pi}{2} \] \(\displaystyle \sin x\) is increasing on \(\displaystyle \left(0,\frac{\pi}{2}\right)\), so \[\sin\frac{\pi}{180}<\sin 1 \]

  10. Exercise 40

    If tan⁡α=mm+1,tan⁡β=12m+1\displaystyle \tan \alpha=\frac{m}{m+1}, \tan \beta=\frac{1}{2 m+1}, then α+β\displaystyle \alpha+\beta is equal to
    (A)
    π2\displaystyle \frac{\pi}{2}
    (B)
    π3\displaystyle \frac{\pi}{3}
    (C)
    π6\displaystyle \frac{\pi}{6}
    (D)
    π4\displaystyle \frac{\pi}{4}

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    NCERT’s answer
    D
    (D) \(\displaystyle \frac{\pi}{4}\). \[\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\dfrac{m}{m+1}+\dfrac{1}{2m+1}}{1-\dfrac{m}{(m+1)(2m+1)}} \] \[=\frac{m(2m+1)+(m+1)}{(m+1)(2m+1)-m}=\frac{2m^2+2m+1}{2m^2+2m+1}=1 \] \[\alpha+\beta=\frac{\pi}{4} \] For \(\displaystyle m>0\), \(\displaystyle \tan\alpha,\tan\beta<1\), so \(\displaystyle 0<\alpha+\beta<\frac{\pi}{2}\).