ICSE Class 10 Trigonometric Identities — Mock Test (2027)
Free online mock test for Trigonometric Identities (ICSE Class 10 Mathematics) — 20 competency-based questions based on the latest CISCE 2027 syllabus, with instant marking. Try the samples below, then take the full test free.
What to expect: This mock test covers key concepts from the Trigonometric Identities chapter — including application-based and competency-focused questions aligned with how ICSE actually sets the paper.
Tip: Attempt without notes first to identify gaps, then review explanations for any wrong answers. Retake after a few days for best retention.
Sample questions
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1.Which of the following correctly proves the identity: $\frac{\cos A}{1 - \sin A} = \sec A + \tan A$?
- A.$\sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A} = \frac{(1 + \sin A)(1 - \sin A)}{\cos A (1 - \sin A)} = \frac{\cos A}{1 - \sin A}$
- B.$\sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A} = \frac{1 + \sin A}{1 - \sin A}$ by cross-multiplying terms
- C.$\frac{\cos A}{1 - \sin A} = \frac{\cos A (1 + \sin A)}{(1 - \sin A)(1 + \sin A)} = \frac{\cos A (1 + \sin A)}{\cos^2 A} = 1 + \sin A$
- D.$\sec A + \tan A = \frac{1 + \sin A}{1 - \sin A}$ by directly combining the terms without rationalization
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2.Prove the trigonometric identity by selecting the correct simplification of the expression: $(1 + \cos A)(1 - \cos A)(1 + \cot^2 A) = 1$. Which of the following steps correctly simplifies the left-hand side to equal the right-hand side?
- A.$(1 - \cos^2 A)(1 + \cot^2 A) = \sin^2 A \times \csc^2 A = 1$
- B.$(1 - \cos^2 A)(1 + \tan^2 A) = \sin^2 A \times \sec^2 A = 1$
- C.$(1 + \cos^2 A)(1 - \cot^2 A) = (1 - \cot^4 A) = 1$
- D.$(1 - \cos A)^2 (1 + \cot^2 A) = (1 - 2\cos A + \cos^2 A) \csc^2 A = 1$
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3.For which condition does the trigonometric identity $\csc^2 A + \csc^2 B = \csc^2 A \csc^2 B$ hold true?
- A.$A + B = 90^\circ$
- B.$A = B$
- C.$A + B = 180^\circ$
- D.$A - B = 90^\circ$
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4.Which of the following correctly proves the identity: $\sec^2 A \cdot \csc^2 A = \tan^2 A + \cot^2 A + 2$?
- A.$\sec^2 A \cdot \csc^2 A = \frac{1}{\cos^2 A \sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\cos^2 A \sin^2 A} = \sec^2 A + \csc^2 A = 2 + \tan^2 A + \cot^2 A$
- B.$\sec^2 A \cdot \csc^2 A = (1 + \tan^2 A)(1 + \cot^2 A) = 1 + \tan^2 A + \cot^2 A + \tan^2 A \cot^2 A$
- C.$\sec^2 A \cdot \csc^2 A = \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \tan^2 A + \cot^2 A$
- D.$\sec^2 A \cdot \csc^2 A = \tan^2 A \cdot \cot^2 A + 2 = 1 + 2 = 3$
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5.Prove that $(\tan A + \cot A)^{2} = \sec^{2} A \csc^{2} A$. Which of the following correctly represents the simplified form of the left-hand side (L.H.S.) before reaching the final identity?
- A.$\frac{1}{\sin^{2} A \cos^{2} A}$
- B.$\frac{\sin^{2} A + \cos^{2} A}{\cos^{2} A \sin^{2} A}$
- C.$\tan^{2} A + \cot^{2} A + 2$
- D.$\frac{1}{\sin A + \cos A}$
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