Question:

Find the value of \( \sin^2 \theta + \cos^2 \theta \): 

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The Pythagoras theorem is the foundation of the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Always check for right-angle triangles when applying this formula.
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Solution and Explanation

Step 1: In \( \triangle ABC \), \( \angle ABC = 90^\circ \) and \( \angle C = \theta \). From the Pythagoras theorem: \[ AB^2 + BC^2 = AC^2. \] Step 2: Dividing throughout by \( AC^2 \): \[ \frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = \frac{AC^2}{AC^2}. \] Step 3: Simplifying, we get: \[ \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1. \] Step 4: Substituting trigonometric ratios: \[ \sin \theta = \frac{BC}{AC}, \quad \cos \theta = \frac{AB}{AC}. \] Thus: \[ \sin^2 \theta + \cos^2 \theta = 1. \]
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