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NCERT Exemplar · Class 11 Mathematics Trigonometric Functions

76 questions · 76 still being checked

EXERCISE 3.3 51–60 (part 6 of 8)

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

    Exercise 51

    If α+β=π4\displaystyle \alpha+\beta=\frac{\pi}{4}, then the value of (1+tan⁡α)(1+tan⁡β)\displaystyle (1+\tan \alpha)(1+\tan \beta) is
    (A)
    1\displaystyle 1 (B) 2\displaystyle 2 (C) - 2\displaystyle 2 (D) Not defined

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    NCERT’s answer
    B
    (B) \(\displaystyle 2\)\[\tan(\alpha+\beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \tan\frac{\pi}{4} = 1 \]\[\tan\alpha + \tan\beta = 1 - \tan\alpha\tan\beta \]\[(1+\tan\alpha)(1+\tan\beta) = 1 + (\tan\alpha + \tan\beta) + \tan\alpha\tan\beta = 1 + (1 - \tan\alpha\tan\beta) + \tan\alpha\tan\beta = 2 \]

  2. Exercise 52

    If sin⁡θ=−45\displaystyle \sin \theta=\frac{-4}{5} and θ\displaystyle \theta lies in third quadrant then the value of cos⁡θ2\displaystyle \cos \frac{\theta}{2} is
    (A)
    15\displaystyle \frac{1}{5}
    (B)
    −110\displaystyle -\frac{1}{\sqrt{10}}
    (C)
    −15\displaystyle -\frac{1}{\sqrt{5}}
    (D)
    110\displaystyle \frac{1}{\sqrt{10}}

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    NCERT’s answer
    C
    (C) \(\displaystyle -\dfrac{1}{\sqrt{5}}\)\[\cos\theta = -\sqrt{1 - \frac{16}{25}} = -\frac{3}{5} \quad (\theta \text{ in third quadrant}) \]\[\cos^2\frac{\theta}{2} = \frac{1+\cos\theta}{2} = \frac{1 - \dfrac{3}{5}}{2} = \frac{1}{5} \]\[\pi < \theta < \frac{3\pi}{2} \;\Longrightarrow\; \frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4} \]\(\displaystyle \dfrac{\theta}{2}\) is in the second quadrant, so cosine is negative.\[\cos\frac{\theta}{2} = -\frac{1}{\sqrt{5}} \]

  3. Exercise 53

    Number of solutions of the equation tan⁡x+sec⁡x=2cos⁡x\displaystyle \tan x+\sec x=2 \cos x lying in the interval [0,2π]\displaystyle [0,2 \pi] is
    (A)
    0\displaystyle 0 (B) 1\displaystyle 1 (C) 2\displaystyle 2 (D) 3\displaystyle 3

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    NCERT’s answer
    C
    (C) \(\displaystyle 2\)\[\frac{\sin x + 1}{\cos x} = 2\cos x \quad (\cos x \neq 0) \]\[\sin x + 1 = 2 - 2\sin^2 x \]\[2\sin^2 x + \sin x - 1 = 0 \]\[(2\sin x - 1)(\sin x + 1) = 0 \]\(\displaystyle \sin x = -1\) gives \(\displaystyle x = \dfrac{3\pi}{2}\), where \(\displaystyle \cos x = 0\), so it is rejected.\[\sin x = \frac{1}{2} \;\Rightarrow\; x = \frac{\pi}{6},\ \frac{5\pi}{6} \]

  4. Exercise 54

    The value of sin⁡π18+sin⁡π9+sin⁡2π9+sin⁡5π18\displaystyle \sin \frac{\pi}{18}+\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\sin \frac{5 \pi}{18} is given by
    (A)
    sin⁡7π18+sin⁡4π9\displaystyle \sin \frac{7 \pi}{18}+\sin \frac{4 \pi}{9}
    (B)
    1\displaystyle 1 (C) cos⁡π6+cos⁡3π7\displaystyle \cos \frac{\pi}{6}+\cos \frac{3 \pi}{7}
    (D)
    cos⁡π9+sin⁡π9\displaystyle \cos \frac{\pi}{9}+\sin \frac{\pi}{9}

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    NCERT’s answer
    A
    (A) \(\displaystyle \sin\dfrac{7\pi}{18} + \sin\dfrac{4\pi}{9}\)\[\sin\frac{\pi}{18} + \sin\frac{5\pi}{18} = 2\sin\frac{\pi}{6}\cos\frac{\pi}{9} = \cos\frac{\pi}{9} = \sin\frac{7\pi}{18} \]\[\sin\frac{\pi}{9} + \sin\frac{2\pi}{9} = 2\sin\frac{\pi}{6}\cos\frac{\pi}{18} = \cos\frac{\pi}{18} = \sin\frac{4\pi}{9} \]Adding the two lines gives the required sum.

  5. Exercise 55

    If A lies in the second quadrant and 3tan⁡ A+4=0\displaystyle 3 \tan \mathrm{~A}+4=0, then the value of 2cot⁡ A−5cos⁡ A+sin⁡A\displaystyle 2 \cot \mathrm{~A}-5 \cos \mathrm{~A}+\sin \mathrm{A} is equal to
    (A)
    −5310\displaystyle \frac{-53}{10}
    (B)
    2310\displaystyle \frac{23}{10}
    (C)
    3710\displaystyle \frac{37}{10}
    (D)
    710\displaystyle \frac{7}{10}

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    NCERT’s answer
    B
    (B) \(\displaystyle \dfrac{23}{10}\)\[\tan A = -\frac{4}{3},\quad A \text{ in second quadrant} \]\[\sin A = \frac{4}{5},\quad \cos A = -\frac{3}{5},\quad \cot A = -\frac{3}{4} \]\[2\cot A - 5\cos A + \sin A = -\frac{3}{2} + 3 + \frac{4}{5} = \frac{23}{10} \]

  6. Exercise 56

    The value of cos⁡248∘−sin⁡212∘\displaystyle \cos ^2 48^{\circ}-\sin ^2 12^{\circ} is
    (A)
    5+18\displaystyle \frac{\sqrt{5}+1}{8}
    (B)
    5−18\displaystyle \frac{\sqrt{5}-1}{8}
    (C)
    5+15\displaystyle \frac{\sqrt{5}+1}{5}
    (D)
    5+122\displaystyle \frac{\sqrt{5}+1}{2 \sqrt{2}}
    [Hint: Use cos⁡2 A−sin⁡2 B=cos⁡(A+B)cos⁡(A−B)\displaystyle \cos ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}=\cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})]

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{\sqrt{5}+1}{8}\)\[\cos^2 48^\circ - \sin^2 12^\circ = \cos(48^\circ + 12^\circ)\cos(48^\circ - 12^\circ) = \cos 60^\circ \cos 36^\circ \]\[\cos 60^\circ \cos 36^\circ = \frac{1}{2}\cdot\frac{\sqrt{5}+1}{4} = \frac{\sqrt{5}+1}{8} \]

  7. Exercise 57

    If tan⁡α=17,tan⁡β=13\displaystyle \tan \alpha=\frac{1}{7}, \tan \beta=\frac{1}{3}, then cos⁡2α\displaystyle \cos 2 \alpha is equal to
    (A)
    sin⁡2β\displaystyle \sin 2 \beta
    (B)
    sin⁡4β\displaystyle \sin 4 \beta
    (C)
    sin⁡3β\displaystyle \sin 3 \beta
    (D)
    cos⁡2β\displaystyle \cos 2 \beta

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    NCERT’s answer
    B
    (B) \(\displaystyle \sin 4\beta\)\[\cos 2\alpha = \frac{1-\tan^2\alpha}{1+\tan^2\alpha} = \frac{1-\dfrac{1}{49}}{1+\dfrac{1}{49}} = \frac{24}{25} \] \[\sin 2\beta = \frac{2\tan\beta}{1+\tan^2\beta} = \frac{3}{5}, \qquad \cos 2\beta = \frac{1-\tan^2\beta}{1+\tan^2\beta} = \frac{4}{5} \] \[\sin 4\beta = 2\sin 2\beta\cos 2\beta = 2\cdot\frac{3}{5}\cdot\frac{4}{5} = \frac{24}{25} = \cos 2\alpha \]

  8. Exercise 58

    If tan⁡θ=ab\displaystyle \tan \theta=\frac{a}{b}, then bcos⁡2θ+asin⁡2θ\displaystyle b \cos 2 \theta+a \sin 2 \theta is equal to
    (A)
    a\displaystyle a (B) b\displaystyle b (C) ab\displaystyle \frac{a}{b}
    (D)
    None

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    NCERT’s answer
    B
    (B) \(\displaystyle b\)\[\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \frac{b^2-a^2}{a^2+b^2}, \qquad \sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2ab}{a^2+b^2} \] \[b\cos 2\theta + a\sin 2\theta = \frac{b(b^2-a^2)+2a^2b}{a^2+b^2} = \frac{b(a^2+b^2)}{a^2+b^2} = b \]

  9. Exercise 59

    If for real values of x,cos⁡θ=x+1x\displaystyle x, \cos \theta=x+\frac{1}{x}, then
    (A)
    θ\displaystyle \theta is an acute angle
    (B)
    θ\displaystyle \theta is right angle
    (C)
    θ\displaystyle \theta is an obtuse angle
    (D)
    No value of θ\displaystyle \theta is possible

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    NCERT’s answer
    D
    (D) No value of \(\displaystyle \theta\) is possible.\(\displaystyle x\) and \(\displaystyle \frac{1}{x}\) have the same sign, so \[\left|x+\frac{1}{x}\right| = |x| + \frac{1}{|x|} \ge 2 \quad \text{(AM-GM)} \] \[|\cos\theta| \le 1 \] The two conditions cannot hold together.
  10. Fill in the blanks in Exercises $\displaystyle 60$ to $\displaystyle 67$ :

    Exercise 60

    The value of sin⁡50∘sin⁡130∘\displaystyle \frac{\sin 50^{\circ}}{\sin 130^{\circ}} is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    $\displaystyle 1$
    \(\displaystyle 1\)\[\sin 130^\circ = \sin(180^\circ-50^\circ) = \sin 50^\circ \] \[\frac{\sin 50^\circ}{\sin 130^\circ} = 1 \]Answer: \(\displaystyle 1\)