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.

Sample questions

1. 1.Prove the trigonometric identity by selecting the correct simplification of the right-hand side: $\frac{1 - \cos A}{1 + \cos A} = (\cot A - \csc A)^{2}$. Which of the following steps correctly simplifies the right-hand side to match the left-hand side?
   • A.$(\cot A - \csc A)^{2} = \frac{(1 - \cos A)^{2}}{1 - \cos^{2} A} = \frac{1 - \cos A}{1 + \cos A}$
   • B.$(\cot A - \csc A)^{2} = \frac{(\cos A - 1)^{2}}{\sin^{2} A} = \frac{1 + \cos A}{1 - \cos A}$
   • C.$(\cot A - \csc A)^{2} = \frac{1 - \cos^{2} A}{(1 + \cos A)^{2}} = \sin^{2} A$
   • D.$(\cot A - \csc A)^{2} = \frac{(\cos A + 1)^{2}}{\sin^{2} A} = \frac{1 + \cos A}{1 - \cos A}$
2. 2.$$\text{(i) } \frac{\sin 80^{\circ}}{\cos 10^{\circ}} + \sin 59^{\circ} \sec 31^{\circ} \\ \text{(ii) } \left(\frac{\sin 39^{\circ}}{\cos 51^{\circ}}\right)^2 + \left(\frac{\cos 51^{\circ}}{\sin 39^{\circ}}\right)^2 \\ \text{(iii) } \frac{\sec 17^{\circ}}{\csc 73^{\circ}} + \frac{\tan 68^{\circ}}{\cot 22^{\circ}} + \cos^2 44^{\circ} + \cos^2 46^{\circ}$$
3. 3.Prove that $\frac{\cos^{3} A + \sin^{3} A}{\cos A + \sin A} + \frac{\cos^{3} A - \sin^{3} A}{\cos A - \sin A} = 2$. What is the simplified value of the given expression?
   • A.$2$
   • B.$1 - 2 \sin A \cos A$
   • C.$2 \cos A \sin A$
   • D.$\cos A + \sin A$
4. 4.If $x = a \sec A \cos B$, $y = b \sec A \sin B$ and $z = c \tan A$, then which of the following identities is true?
   • A.$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$
   • B.$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
   • C.$\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$
   • D.$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z^2}{c^2}$
5. 5.Which of the following correctly proves the identity: $\tan^2 A - \tan^2 B = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cdot \cos^2 B}$?
   • A.Express $\tan^2 A$ and $\tan^2 B$ in terms of $\sin$ and $\cos$, combine into a single fraction, and simplify using $\sin^2 x + \cos^2 x = 1$.
   • B.Rewrite $\tan^2 A - \tan^2 B$ as $(\tan A - \tan B)^2$ and expand using algebraic identities.
   • C.Convert $\sin^2 A - \sin^2 B$ to $\cos^2 B - \cos^2 A$ and factor using difference of squares.
   • D.Use the identity $\tan^2 x = \sec^2 x - 1$ and simplify the expression directly without converting to $\sin$ and $\cos$.