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 trigonometric identity by selecting the expression that correctly simplifies the left-hand side to match the right-hand side: $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = \sec A \csc A + 1$. Which of the following represents the correct simplification of the left-hand side?
- A.$\frac{1 + \sin A \cos A}{\sin A \cos A}$
- B.$\frac{\sin A + \cos A}{\sin A - \cos A}$
- C.$\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}$
- D.$\frac{\sin^3 A + \cos^3 A}{\sin A \cos A (\sin A - \cos A)}$
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2.(ii) Using the table, find the value of: $\cos 23^\circ 6'$
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3.Which of the following correctly simplifies the expression $\sin A (1 + \tan A) + \cos A (1 + \cot A)$?
- A.$\sec A + \csc A$
- B.$\sin A + \cos A$
- C.$\frac{1 + \sin A + \cos A}{\cos A \sin A}$
- D.$\tan A + \cot A$
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4.If $x = p\sec \theta$ and $y = q\tan \theta$, then
- a.$x^{2} - y^{2} = p^{2}q^{2}$
- b.$x^{2}q^{2} - y^{2}p^{2} = pq$
- c.$x^{2}q^{2} - y^{2}p^{2} = \frac{1}{p^{2}q^{2}}$
- d.$x^{2}q^{2} - y^{2}p^{2} = p^{2}q^{2}$
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5.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.$\frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{\cos^2 A + \sin^2 A}{\cos^2 A \sin^2 A} = \frac{1}{\cos^2 A \sin^2 A} + 1$
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