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Chapter 8: Introduction to Trigonometry

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This page provides comprehensive Chapter 8: Introduction to Trigonometry - Standard Worksheet - SJMaths. Standard level practice worksheet for Class 10 Introduction to Trigonometry. Practice for CBSE Board Exams.

Standard Level Worksheet

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  1. Question 1: Prove that $\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta} = \tan \theta$.
    Solution:
    Step 1: LHS $= \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)}$.
    Step 2: Use $\sin^2 \theta = 1 - \cos^2 \theta$ in numerator: $1 - 2(1-\cos^2 \theta) = 2\cos^2 \theta - 1$.
    Step 3: LHS $= \frac{\sin \theta(2\cos^2 \theta - 1)}{\cos \theta(2\cos^2 \theta - 1)} = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
  2. Question 2: Evaluate: $2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$.
    Solution:
    Step 1: Substitute values: $2(1)^2 + (\frac{\sqrt{3}}{2})^2 - (\frac{\sqrt{3}}{2})^2$.
    Step 2: $2 + \frac{3}{4} - \frac{3}{4} = 2$.
  3. Question 3: If $\tan(A+B) = \sqrt{3}$ and $\tan(A-B) = \frac{1}{\sqrt{3}}$; $0^\circ < A+B \le 90^\circ; A > B$, find $A$ and $B$.
    Solution:
    Step 1: $A+B = 60^\circ$ and $A-B = 30^\circ$.
    Step 2: Add equations: $2A = 90^\circ \Rightarrow A = 45^\circ$.
    Step 3: Substitute $A$: $45^\circ + B = 60^\circ \Rightarrow B = 15^\circ$.
  4. Question 4: Prove that $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$.
    Solution:
    Step 1: Expand: $(\sin^2 A + \csc^2 A + 2) + (\cos^2 A + \sec^2 A + 2)$.
    Step 2: Group: $(\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 4$.
    Step 3: $1 + (1+\cot^2 A) + (1+\tan^2 A) + 4 = 7 + \tan^2 A + \cot^2 A$.
  5. Question 5: If $\sin(A-B) = \frac{1}{2}, \cos(A+B) = \frac{1}{2}$, find A and B.
    Solution:
    Step 1: $A-B = 30^\circ$ and $A+B = 60^\circ$.
    Step 2: Add equations: $2A = 90^\circ \Rightarrow A = 45^\circ$.
    Step 3: Substitute $A$: $45^\circ - B = 30^\circ \Rightarrow B = 15^\circ$.
  6. Question 6: Prove that $\frac{\cos A}{1+\sin A} + \frac{1+\sin A}{\cos A} = 2\sec A$.
    Solution:
    Step 1: LHS $= \frac{\cos^2 A + (1+\sin A)^2}{(1+\sin A)\cos A}$.
    Step 2: Numerator $= \cos^2 A + 1 + 2\sin A + \sin^2 A = 2 + 2\sin A = 2(1+\sin A)$.
    Step 3: $\frac{2(1+\sin A)}{(1+\sin A)\cos A} = \frac{2}{\cos A} = 2\sec A$.
  7. Question 7: Prove that $\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$.
    Solution:
    Step 1: Multiply numerator and denominator by $(1+\sin A)$ inside the root.
    Step 2: $\sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}$.
    Step 3: $\frac{1+\sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A$.
  8. Question 8: If $\cot \theta = \frac{7}{8}$, evaluate $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$.
    Solution:
    Step 1: Simplify expression: $\frac{1-\sin^2 \theta}{1-\cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta$.
    Step 2: Substitute value: $(\frac{7}{8})^2 = \frac{49}{64}$.
  9. Question 9: Prove that $\frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta} = 1 + \sec \theta \csc \theta$.
    Solution:
    Step 1: Convert to sin, cos: $\frac{\sin/\cos}{1-\cos/\sin} + \frac{\cos/\sin}{1-\sin/\cos} = \frac{\sin^2}{\cos(\sin-\cos)} - \frac{\cos^2}{\sin(\sin-\cos)}$.
    Step 2: $\frac{\sin^3 - \cos^3}{\sin\cos(\sin-\cos)} = \frac{(\sin-\cos)(\sin^2+\cos^2+\sin\cos)}{\sin\cos(\sin-\cos)}$.
    Step 3: $\frac{1+\sin\cos}{\sin\cos} = \frac{1}{\sin\cos} + 1 = 1 + \sec\theta\csc\theta$.
  10. Question 10: If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, show that $m^2 - n^2 = 4\sqrt{mn}$.
    Solution:
    Step 1: LHS $= m^2 - n^2 = (m-n)(m+n) = (2\sin\theta)(2\tan\theta) = 4\tan\theta\sin\theta$.
    Step 2: RHS $= 4\sqrt{mn} = 4\sqrt{(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)} = 4\sqrt{\tan^2\theta - \sin^2\theta}$.
    Step 3: $4\sqrt{\frac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta} = 4\sqrt{\sin^2\theta(\frac{1}{\cos^2\theta}-1)} = 4\sqrt{\sin^2\theta\tan^2\theta} = 4\sin\theta\tan\theta$.
    Step 4: LHS = RHS.
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