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Chapter 8: Introduction to Trigonometry

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This page provides comprehensive Chapter 8: Introduction to Trigonometry - MCQ Worksheet - SJMaths. Multiple Choice Questions (MCQ) worksheet for Class 10 Introduction to Trigonometry. Practice for CBSE Board Exams.

MCQ Worksheet

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  1. Question 1: If $\sin A = \frac{3}{4}$, then $\cos A$ is equal to:
    (A) $\frac{4}{3}$
    (B) $\frac{\sqrt{7}}{4}$
    (C) $\frac{3}{\sqrt{7}}$
    (D) $\frac{\sqrt{7}}{3}$
    Solution: (B) $\frac{\sqrt{7}}{4}$
    Step 1: $\cos^2 A = 1 - \sin^2 A = 1 - (\frac{3}{4})^2 = 1 - \frac{9}{16} = \frac{7}{16}$.
    Step 2: $\cos A = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.
  2. Question 2: The value of $\tan 45^\circ + \cot 45^\circ$ is:
    (A) 1
    (B) 2
    (C) 0
    (D) $\sqrt{2}$
    Solution: (B) 2
    Step 1: $\tan 45^\circ = 1$ and $\cot 45^\circ = 1$.
    Step 2: $1 + 1 = 2$.
  3. Question 3: If $\cos A = \frac{4}{5}$, then the value of $\tan A$ is:
    (A) $\frac{3}{5}$
    (B) $\frac{3}{4}$
    (C) $\frac{4}{3}$
    (D) $\frac{5}{3}$
    Solution: (B) $\frac{3}{4}$
    Step 1: In a right triangle, if base=4 and hypotenuse=5, then perpendicular = $\sqrt{5^2 - 4^2} = \sqrt{25-16} = 3$.
    Step 2: $\tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4}$.
  4. Question 4: The value of $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}$ is:
    (A) $\tan 90^\circ$
    (B) 1
    (C) $\sin 45^\circ$
    (D) 0
    Solution: (D) 0
    Step 1: $\tan 45^\circ = 1$.
    Step 2: $\frac{1 - 1^2}{1 + 1^2} = \frac{0}{2} = 0$.
  5. Question 5: $9 \sec^2 A - 9 \tan^2 A$ is equal to:
    (A) 1
    (B) 9
    (C) 8
    (D) 0
    Solution: (B) 9
    Step 1: Factor out 9: $9(\sec^2 A - \tan^2 A)$.
    Step 2: Using identity $\sec^2 A - \tan^2 A = 1$, we get $9(1) = 9$.
  6. Question 6: If $\sin \theta = \cos \theta$, then the value of $\theta$ is:
    (A) $0^\circ$
    (B) $30^\circ$
    (C) $45^\circ$
    (D) $90^\circ$
    Solution: (C) $45^\circ$
    Step 1: $\frac{\sin \theta}{\cos \theta} = 1 \Rightarrow \tan \theta = 1$.
    Step 2: $\tan 45^\circ = 1$, so $\theta = 45^\circ$.
  7. Question 7: The value of $(\sin 30^\circ + \cos 30^\circ) - (\sin 60^\circ + \cos 60^\circ)$ is:
    (A) -1
    (B) 0
    (C) 1
    (D) 2
    Solution: (B) 0
    Step 1: $(\frac{1}{2} + \frac{\sqrt{3}}{2}) - (\frac{\sqrt{3}}{2} + \frac{1}{2})$.
    Step 2: Both terms are identical, so the difference is 0.
  8. Question 8: If $\Delta ABC$ is right angled at $C$, then the value of $\cos(A+B)$ is:
    (A) 0
    (B) 1
    (C) $\frac{1}{2}$
    (D) $\frac{\sqrt{3}}{2}$
    Solution: (A) 0
    Step 1: In $\Delta ABC$, $A+B+C = 180^\circ$. Since $C=90^\circ$, $A+B = 90^\circ$.
    Step 2: $\cos(A+B) = \cos(90^\circ) = 0$.
  9. Question 9: Given that $\sin \alpha = \frac{1}{2}$ and $\cos \beta = \frac{1}{2}$, then the value of $(\alpha + \beta)$ is:
    (A) $0^\circ$
    (B) $30^\circ$
    (C) $60^\circ$
    (D) $90^\circ$
    Solution: (D) $90^\circ$
    Step 1: $\sin \alpha = \frac{1}{2} \Rightarrow \alpha = 30^\circ$.
    Step 2: $\cos \beta = \frac{1}{2} \Rightarrow \beta = 60^\circ$.
    Step 3: $\alpha + \beta = 30^\circ + 60^\circ = 90^\circ$.
  10. Question 10: $(1 + \tan^2 A)(1 + \sin A)(1 - \sin A)$ is equal to:
    (A) 0
    (B) 1
    (C) 2
    (D) $\cos^2 A$
    Solution: (B) 1
    Step 1: $(1 + \tan^2 A) = \sec^2 A$.
    Step 2: $(1 + \sin A)(1 - \sin A) = 1 - \sin^2 A = \cos^2 A$.
    Step 3: $\sec^2 A \cdot \cos^2 A = \frac{1}{\cos^2 A} \cdot \cos^2 A = 1$.
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