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: $\frac{\sin A \tan A}{1 - \cos A} = 1 + \sec A$?
- A.$\frac{\sin A \tan A}{1 - \cos A} = \frac{\sin^2 A}{\cos A (1 - \cos A)} = \frac{1 - \cos^2 A}{\cos A (1 - \cos A)} = \frac{1 + \cos A}{\cos A} = 1 + \sec A$
- B.$\frac{\sin A \tan A}{1 - \cos A} = \frac{\sin A \cdot \frac{\sin A}{\cos A}}{1 - \cos A} = \frac{\sin^2 A}{1 - \cos A} = 1 - \cos A = 1 - \sec A$
- C.$\frac{\sin A \tan A}{1 - \cos A} = \frac{\sin A \cdot \frac{1}{\cos A}}{1 - \cos A} = \frac{\tan A}{1 - \cos A} = \frac{\sec A}{1 - \cos A} = \sec A$
- D.$\frac{\sin A \tan A}{1 - \cos A} = \frac{\sin^2 A}{\cos A (1 - \cos A)} = \frac{1 - \cos^2 A}{1 - \cos A} = 1 + \cos A$
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2.Find the value of $x$, if: $\cos x = \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ$.
- A.$x = 30^\circ$
- B.$x = 60^\circ$
- C.$x = 45^\circ$
- D.$x = 90^\circ$
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3.If $\cos \theta = 0.51$, find: (i) θ (ii) $\sin \theta$ (iii) $\tan \theta$
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4.Which of the following identities is equivalent to the given expression: $\cos A (1 + \cot A) + \sin A (1 + \tan A)$?
- A.$\sec A + \csc A$
- B.$\sin A + \cos A$
- C.$\frac{1}{\sin A \cos A}$
- D.$\tan A + \cot A$
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5.Which of the following correctly proves the identity $\tan^2 A \sec^2 B - \sec^2 A \tan^2 B = \tan^2 A - \tan^2 B$?
- A.Substitute $\sec^2 B = 1 + \tan^2 B$ and $\sec^2 A = 1 + \tan^2 A$ into the LHS and simplify to match the RHS.
- B.Factor out $\tan^2 A \tan^2 B$ from the LHS and simplify using $\sec^2 A = 1 + \tan^2 A$.
- C.Rewrite $\tan^2 A$ and $\tan^2 B$ in terms of $\sin$ and $\cos$, then cross-multiply to cancel terms.
- D.Use $\sec^2 A = \tan^2 A + 1$ only on the first term of the LHS and ignore the second term.
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