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NCERT Exemplar · Class 10 Mathematics Introduction to Trigonometry and its Applications

60 questions · 60 still being checked

EXERCISE 8.1 1–10 (part 1 of 7)

  1. Choose the correct answer from the given four options:

    Exercise 1

    If cos⁡A=45\displaystyle \cos \mathrm{A}=\frac{4}{5}, then the value of tan⁡A\displaystyle \tan \mathrm{A} is
    (A)
    35\displaystyle \frac{3}{5}
    (B)
    34\displaystyle \frac{3}{4}
    (C)
    43\displaystyle \frac{4}{3}
    (D)
    53\displaystyle \frac{5}{3}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle \frac{3}{4}\)Take adjacent \(\displaystyle =4\), hypotenuse \(\displaystyle =5\) for \(\displaystyle \cos A=\frac{4}{5}\). \[\text{opposite}=\sqrt{5^{2}-4^{2}}=3 \] \[\tan A=\frac{3}{4} \]

  2. Exercise 2

    If sin⁡A=12\displaystyle \sin \mathrm{A}=\frac{1}{2}, then the value of cot⁡A\displaystyle \cot \mathrm{A} is
    (A)
    3\displaystyle \sqrt{3}
    (B)
    13\displaystyle \frac{1}{\sqrt{3}}
    (C)
    32\displaystyle \frac{\sqrt{3}}{2}

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    NCERT’s answer
    (A)
    (A) \(\displaystyle \sqrt{3}\)\(\displaystyle \sin A=\frac{1}{2}\) gives \(\displaystyle A=30^{\circ}\). \[\cot A=\cot 30^{\circ}=\sqrt{3} \]

  3. Exercise 3

    The value of the expression [cosec⁡(75∘+θ)−sec⁡(15∘−θ)−tan⁡(55∘+θ)+cot⁡(35∘−θ)]\displaystyle \left[\operatorname{cosec}\left(75^{\circ}+\theta\right)-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\cot \left(35^{\circ}-\theta\right)\right] is
    (A)
    −1\displaystyle -1 (B) 0\displaystyle 0 (C) 1\displaystyle 1 (D) 32\displaystyle \frac{3}{2}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 0\)\(\displaystyle (75^{\circ}+\theta)+(15^{\circ}-\theta)=90^{\circ}\), so \[\sec(15^{\circ}-\theta)=\csc(75^{\circ}+\theta) \] \(\displaystyle (55^{\circ}+\theta)+(35^{\circ}-\theta)=90^{\circ}\), so \[\cot(35^{\circ}-\theta)=\tan(55^{\circ}+\theta) \] Both pairs cancel.

  4. Exercise 4

    Given that sin⁡θ=ab\displaystyle \sin \theta=\frac{a}{b}, then cos⁡θ\displaystyle \cos \theta is equal to
    (A)
    bb2−a2\displaystyle \frac{b}{\sqrt{b^2-a^2}}
    (B)
    ba\displaystyle \frac{b}{a}
    (C)
    b2−a2b\displaystyle \frac{\sqrt{b^2-a^2}}{b}
    (D)
    ab2−a2\displaystyle \frac{a}{\sqrt{b^2-a^2}}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle \dfrac{\sqrt{b^{2}-a^{2}}}{b}\)\[\sin^{2}\theta+\cos^{2}\theta=1 \] \[\cos\theta=\sqrt{1-\frac{a^{2}}{b^{2}}}=\frac{\sqrt{b^{2}-a^{2}}}{b} \]

  5. Exercise 5

    If cos⁡(α+β)=0\displaystyle \cos (\alpha+\beta)=0, then sin⁡(α−β)\displaystyle \sin (\alpha-\beta) can be reduced to
    (A)
    cos⁡β\displaystyle \cos \beta
    (B)
    cos⁡2β\displaystyle \cos 2 \beta
    (C)
    sin⁡α\displaystyle \sin \alpha
    (D)
    sin⁡2α\displaystyle \sin 2 \alpha

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    NCERT’s answer
    (B)
    (B) \(\displaystyle \cos 2\beta\)\[\cos(\alpha+\beta)=0 \implies \alpha+\beta=90^{\circ} \implies \alpha=90^{\circ}-\beta \] \[\sin(\alpha-\beta)=\sin(90^{\circ}-2\beta)=\cos 2\beta \]

  6. Exercise 6

    The value of (tan⁡1∘tan⁡2∘tan⁡3∘…tan⁡89∘)\displaystyle \left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right) is
    (A)
    0\displaystyle 0 (B) 1\displaystyle 1 (C) 2\displaystyle 2 (D) 12\displaystyle \frac{1}{2}

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 1\)Pair each angle with its complement, \(\displaystyle 1^{\circ}\) with \(\displaystyle 89^{\circ}\) up to \(\displaystyle 44^{\circ}\) with \(\displaystyle 46^{\circ}\), plus \(\displaystyle \tan 45^{\circ}\). \[\tan\theta\,\tan(90^{\circ}-\theta)=\tan\theta\,\cot\theta=1 \] \[\tan1^{\circ}\tan2^{\circ}\cdots\tan89^{\circ}=1^{44}\cdot\tan45^{\circ}=1 \]

  7. Exercise 7

    If cos⁡9α=sin⁡α\displaystyle \cos 9 \alpha=\sin \alpha and 9α<90∘\displaystyle 9 \alpha<90^{\circ}, then the value of tan⁡5α\displaystyle \tan 5 \alpha is
    (A)
    13\displaystyle \frac{1}{\sqrt{3}}
    (B)
    3\displaystyle \sqrt{3}
    (C)
    1\displaystyle 1 (D) 0\displaystyle 0

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 1\)\[\cos9\alpha=\sin\alpha=\cos(90^{\circ}-\alpha) \] \[9\alpha=90^{\circ}-\alpha \implies \alpha=9^{\circ} \] \[\tan5\alpha=\tan45^{\circ}=1 \]

  8. Exercise 8

    If ΔABC\displaystyle \Delta \mathrm{ABC} is right angled at C, then the value of cos⁡(A+B)\displaystyle \cos (\mathrm{A}+\mathrm{B}) is
    (A)
    0\displaystyle 0 (B) 1\displaystyle 1 (C) 12\displaystyle \frac{1}{2}
    (D)
    32\displaystyle \frac{\sqrt{3}}{2}

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 0\)Right angle at \(\displaystyle C\) gives \[A+B=90^{\circ} \] \[\cos(A+B)=\cos90^{\circ}=0 \]

  9. Exercise 9

    If sin⁡A+sin⁡2A=1\displaystyle \sin \mathrm{A}+\sin ^2 \mathrm{A}=1, then the value of the expression (cos⁡2A+cos⁡4A)\displaystyle \left(\cos ^2 \mathrm{A}+\cos ^4 \mathrm{A}\right) is
    (A)
    1\displaystyle 1 (B) 12\displaystyle \frac{1}{2}
    (C)
    2\displaystyle 2 (D) 3\displaystyle 3

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 1\) \[\sin A + \sin^2 A = 1 \] \[\sin A = 1-\sin^2A = \cos^2 A \] \[\cos^2 A + \cos^4 A = \cos^2A+(\cos^2A)^2 = \sin A+\sin^2A = 1 \]

  10. Exercise 10

    Given that sin⁡α=12\displaystyle \sin \alpha=\frac{1}{2} and cos⁡β=12\displaystyle \cos \beta=\frac{1}{2}, then the value of (α+β)\displaystyle (\alpha+\beta) is
    (A)
    0\displaystyle 0° (B) 30\displaystyle 30°
    (C)
    60\displaystyle 60°
    (D)
    90\displaystyle 90°

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 90^\circ\) \[\sin\alpha=\tfrac12 \implies \alpha=30^\circ \] \[\cos\beta=\tfrac12 \implies \beta=60^\circ \] \[\alpha+\beta = 30^\circ+60^\circ = 90^\circ \]