1. **Prove the trigonometric identity:** Given: $$(\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta$$ Step 1: Expand the left side using the formula $$(a+b)^2 = a^2 + 2ab + b^2$$ $$ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta $$ Step 2: Use the Pythagorean identity $$\sin^2 \theta + \cos^2 \theta = 1$$ So, $$ \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta = 1 + 2 \sin \theta \cos \theta $$ This matches the right side, so the identity is proven. 2. **Find $h$ such that points $A(\sqrt{3}, -1)$, $B(0, 2)$, and $C(h, -2)$ form a right triangle with right angle at $A$.** Step 1: Calculate vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$: $$ \overrightarrow{AB} = (0 - \sqrt{3}, 2 - (-1)) = (-\sqrt{3}, 3) $$ $$ \overrightarrow{AC} = (h - \sqrt{3}, -2 - (-1)) = (h - \sqrt{3}, -1) $$ Step 2: For a right angle at $A$, vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ must be perpendicular, so their dot product is zero: $$ \overrightarrow{AB} \cdot \overrightarrow{AC} = 0 $$ Calculate the dot product: $$ (-\sqrt{3})(h - \sqrt{3}) + 3(-1) = 0 $$ Step 3: Simplify: $$ -\sqrt{3}h + 3 - 3 = 0 $$ $$ -\sqrt{3}h = 0 $$ Step 4: Solve for $h$: $$ h = 0 $$ 3. **Show that triangle with vertices $A(6,1)$, $B(2,7)$, and $C(-6,-7)$ is a right triangle using slopes.** Step 1: Calculate slopes of sides: $$ m_{AB} = \frac{7 - 1}{2 - 6} = \frac{6}{-4} = -\frac{3}{2} $$ $$ m_{BC} = \frac{-7 - 7}{-6 - 2} = \frac{-14}{-8} = \frac{7}{4} $$ $$ m_{AC} = \frac{-7 - 1}{-6 - 6} = \frac{-8}{-12} = \frac{2}{3} $$ Step 2: Check if any two slopes are negative reciprocals (perpendicular): Check $m_{AB} \times m_{AC}$: $$ -\frac{3}{2} \times \frac{2}{3} = -1 $$ Since the product is $-1$, sides $AB$ and $AC$ are perpendicular, so the triangle is right angled at $A$. **Final answers:** 1. Identity proven. 2. $h = 0$ 3. Triangle is right angled at $A$.