ICSE Class 10

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.

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Sample questions

  1. 1.Which of the following correctly proves the identity: $\csc A (1 + \cos A)(\csc A - \cot A) = 1$?
    • A.$(1 + \cos A)(1 - \cos A) = 1 - \cos^2 A = \sin^2 A$, which simplifies to 1 when multiplied by $\csc A$.
    • B.$\csc A (1 + \cos A)(\csc A - \cot A) = (1 + \cos A)(\csc^2 A - \csc A \cot A) = 1 + \cos A - \cos A = 1$.
    • C.$\csc A (1 + \cos A)(\csc A - \cot A) = (1 + \cos A)(1 - \sin A) = 1 - \sin A + \cos A - \sin A \cos A$.
    • D.$\csc A - \cot A = \sin A$, so the expression simplifies to $\csc A (1 + \cos A) \sin A = 1 + \cos A$.
  2. 2.$$\text{(i) } \frac{\cos^2 34^{\circ} + \cos^2 56^{\circ}}{\sin^2 59^{\circ} + \sin^2 31^{\circ}} \\ \text{(ii) } \frac{\csc^2 67^{\circ} - \tan^2 23^{\circ}}{\sec^2 20^{\circ} - \cot^2 70^{\circ}}$$
  3. 3.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.$\tan A - \cot A = \frac{\sin A}{\cos A} - \frac{\cos A}{\sin A} = \frac{2\cos^2 A - 1}{\sin A \cos A}$
  4. 4.Prove the trigonometric identity by selecting the correct simplification step: $\sec^2 A + \operatorname{cosec}^2 A = \sec^2 A \operatorname{cosec}^2 A$
    • A.$\frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A \cos^2 A} = \frac{1}{\sin^2 A \cos^2 A}$
    • B.$\frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{2}{\sin^2 A \cos^2 A}$ (using $\sin^2 A + \cos^2 A = 2$)
    • C.$\sec^2 A + \operatorname{cosec}^2 A = \frac{1}{\cos^2 A + \sin^2 A} = 1$
    • D.$\frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{\cos^2 A + \sin^2 A}{\cos^2 A \sin^2 A} = \frac{1}{\cos^2 A \sin^2 A} + 1$
  5. 5.Which of the following correctly shows that $\sin 42^\circ \sec 48^\circ + \cos 42^\circ \csc 48^\circ = 2$?
    • A.$\sin 42^\circ \sec 48^\circ + \cos 42^\circ \csc 48^\circ = \sin 42^\circ \csc 42^\circ + \cos 42^\circ \sec 42^\circ = 1 + 1 = 2$
    • B.$\sin 42^\circ \sec 48^\circ + \cos 42^\circ \csc 48^\circ = \sin 42^\circ \cos 48^\circ + \cos 42^\circ \sin 48^\circ = \sin 90^\circ = 1$
    • C.$\sin 42^\circ \sec 48^\circ + \cos 42^\circ \csc 48^\circ = \sec 42^\circ \csc 42^\circ = 1/(\sin 42^\circ \cos 42^\circ) = 2$
    • D.$\sin 42^\circ \sec 48^\circ + \cos 42^\circ \csc 48^\circ = \sin^2 42^\circ + \cos^2 42^\circ = 1$

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