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.Prove the following trigonometric identity by selecting the correct simplified form of the left-hand side (L.H.S.): $\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = ?$
- A.$\cos A + \sin A$
- B.$\frac{1}{\cos A - \sin A}$
- C.$\cos^2 A + \sin^2 A$
- D.$\frac{\cos A - \sin A}{\cos A + \sin A}$
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2.Which of the following simplifies to $\tan A$?
- A.$\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}$
- B.$\frac{\sin A + 2 \sin^3 A}{2 \cos^3 A - \cos A}$
- C.$\frac{2 \sin^3 A - \sin A}{\cos A - 2 \cos^3 A}$
- D.$\frac{\sin A - 2 \sin^3 A}{\cos A - 2 \cos^3 A}$
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3.(iv) Using the table, find the value of: $\cos 85^\circ 8'$
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4.Eliminate $\theta$ between the equations: $x = a \sec^{3} \theta$ and $y = b \tan^{3} \theta$. Which of the following equations results from this elimination?
- A.$\left(\frac{x}{a}\right)^{\frac{2}{3}} - \left(\frac{y}{b}\right)^{\frac{2}{3}} = 1$
- B.$\left(\frac{x}{a}\right)^{\frac{1}{3}} + \left(\frac{y}{b}\right)^{\frac{1}{3}} = 1$
- C.$\left(\frac{x}{a}\right)^2 - \left(\frac{y}{b}\right)^2 = 1$
- D.$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
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5.Which of the following correctly proves the identity: $\tan A - \cot A = \frac{1 - 2\cos^2 A}{\sin A \cos A}$?
- A.$\tan A - \cot A = \frac{\sin A}{\cos A} - \frac{\cos A}{\sin A} = \frac{\sin^2 A - \cos^2 A}{\sin A \cos A} = \frac{1 - 2\cos^2 A}{\sin A \cos A}$
- B.$\tan A - \cot A = \frac{\sin A - \cos A}{\cos A \sin A} = \frac{1 - 2\cos^2 A}{\sin A \cos A}$
- C.$\tan A - \cot A = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1 - 2\cos^2 A}{\sin A \cos A}$
- D.$\tan A - \cot A = \frac{\sin A}{\cos A} - \frac{\cos A}{\sin A} = \frac{2\cos^2 A - 1}{\sin A \cos A}$
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