1. **Problem:** Simplify the expression $$\sec^2 x - 1$$ and identify which function it equals. 2. **Recall the Pythagorean identity:** $$\sec^2 x = 1 + \tan^2 x$$. 3. Substitute this into the expression: $$\sec^2 x - 1 = (1 + \tan^2 x) - 1$$ 4. Simplify by canceling 1: $$\sec^2 x - 1 = \cancel{1} + \tan^2 x - \cancel{1} = \tan^2 x$$ 5. **Answer:** $$\sec^2 x - 1 = \tan^2 x$$. --- 1. **Problem:** Simplify $$\frac{1}{\sec x + 1} - \frac{1}{\sec x - 1}$$. 2. Find a common denominator: $$ (\sec x + 1)(\sec x - 1) = \sec^2 x - 1 $$ 3. Write the expression as: $$ \frac{\sec x - 1}{\sec^2 x - 1} - \frac{\sec x + 1}{\sec^2 x - 1} = \frac{(\sec x - 1) - (\sec x + 1)}{\sec^2 x - 1} $$ 4. Simplify numerator: $$ (\sec x - 1) - (\sec x + 1) = \sec x - 1 - \sec x - 1 = -2 $$ 5. So expression is: $$ \frac{-2}{\sec^2 x - 1} $$ 6. Recall $$\sec^2 x - 1 = \tan^2 x$$, so: $$ \frac{-2}{\tan^2 x} = -2 \cot^2 x $$ 7. **Answer:** $$-2 \cot^2 x$$. --- 1. **Problem:** Simplify $$(\cot x + \csc x)(\cot x - \csc x)$$. 2. Use difference of squares formula: $$ (a+b)(a-b) = a^2 - b^2 $$ 3. So: $$ (\cot x)^2 - (\csc x)^2 = \cot^2 x - \csc^2 x $$ 4. Recall identity: $$ \csc^2 x = 1 + \cot^2 x $$ 5. Substitute: $$ \cot^2 x - (1 + \cot^2 x) = \cot^2 x - 1 - \cot^2 x = -1 $$ 6. **Answer:** $$-1$$. --- 1. **Problem:** Simplify $$(6 - 6 \sin x)(6 + 6 \sin x)$$. 2. Use difference of squares: $$ (a - b)(a + b) = a^2 - b^2 $$ 3. Here: $$ a = 6, b = 6 \sin x $$ 4. So: $$ 6^2 - (6 \sin x)^2 = 36 - 36 \sin^2 x $$ 5. Factor out 36: $$ 36(1 - \sin^2 x) $$ 6. Recall identity: $$ 1 - \sin^2 x = \cos^2 x $$ 7. So expression is: $$ 36 \cos^2 x $$ 8. **Answer:** $$36 \cos^2 x$$. --- 1. **Problem:** Simplify $$\tan x - \frac{\sec^2 x}{\tan x}$$. 2. Write as a single fraction: $$ \frac{\tan^2 x - \sec^2 x}{\tan x} $$ 3. Recall identity: $$ \sec^2 x = 1 + \tan^2 x $$ 4. Substitute numerator: $$ \tan^2 x - (1 + \tan^2 x) = \tan^2 x - 1 - \tan^2 x = -1 $$ 5. So expression is: $$ \frac{-1}{\tan x} = - \cot x $$ 6. **Answer:** $$- \cot x$$. --- 1. **Problem:** Rewrite $$\sqrt{625 - 25x^2}$$ as a trigonometric function of $$\theta$$ using substitution $$x = 5 \sin \theta$$, where $$0 < \theta < \frac{\pi}{2}$$. 2. Substitute: $$ \sqrt{625 - 25(5 \sin \theta)^2} = \sqrt{625 - 25 \cdot 25 \sin^2 \theta} $$ 3. Simplify inside the root: $$ 625 - 625 \sin^2 \theta = 625(1 - \sin^2 \theta) $$ 4. Recall identity: $$ 1 - \sin^2 \theta = \cos^2 \theta $$ 5. So: $$ \sqrt{625 \cos^2 \theta} = 25 \cos \theta $$ 6. **Answer:** $$25 \cos \theta$$.