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.

Sample questions

  1. 1.(iii) Using tables find the value of: $\tan 31^\circ 27'$
    • A.0.6084
    • B.0.6104
    • C.0.6116
    • D.0.6146
  2. 2.If $\sin A = \frac{1}{2}$, then the value of $\cot A$ is
    • a.$\sqrt{3}$
    • b.$\frac{1}{\sqrt{3}}$
    • c.$\frac{\sqrt{3}}{2}$
    • d.1
  3. 3.Given A is an acute angle and $13 \sin A = 5$, evaluate without using tables: $\frac{(5 \sin A - 2 \cos A)}{\tan A}$.
  4. 4.Which of the following correctly proves the identity $\frac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta} = \frac{\sec \theta + 1}{\sec \theta - 1}$?
    • A.Express $\tan \theta$ as $\frac{\sin \theta}{\cos \theta}$, factor out $\sin \theta$ from numerator and denominator, then simplify to match the R.H.S.
    • B.Multiply numerator and denominator by $\cos \theta$ directly without expressing $\tan \theta$ in terms of $\sin \theta$ and $\cos \theta$.
    • C.Replace $\tan \theta$ with $\sin \theta$ and $\sec \theta$ with $\cos \theta$, then cross-multiply to verify the identity.
    • D.Square both sides of the equation and use the Pythagorean identity to show equality.
  5. 5.Eliminate θ from the following equations: $x = h + a \cos \theta$, $y = k + b \sin \theta$. Which of the following relations is obtained?
    • A.$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$
    • B.$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$
    • C.$\frac{(x - k)^2}{a^2} + \frac{(y - h)^2}{b^2} = 1$
    • D.$(x - h)^2 + (y - k)^2 = a^2 + b^2$

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