Mathematics — Standard (041) — CBSE Class 10 board question

Given: $\sin A = y$

Using the identity $\sin^2 A + \cos^2 A = 1$:

$$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - y^2}$$

Using $\tan A = \dfrac{\sin A}{\cos A}$:

$$\tan A = \dfrac{y}{\sqrt{1-y^2}}$$

Source: Chapter 8, Section 8.2 & 8.5

---

• The examiner expects you to use the Pythagorean identity $\sin^2A + \cos^2A = 1$ directly to find $\cos A$, then use $\tan A = \frac{\sin A}{\cos A}$.
• Assume angle A is acute, so $\cos A$ is positive — take the positive square root only.
• Both results must be clearly written in terms of $y$; the substitution step should be shown for full marks.