1. **Problem statement:** Prove the identity $ \frac{1}{\sin x} - \sin x = \frac{\cos x}{\tan x} $. 2. **Recall definitions and formulas:** - $ \tan x = \frac{\sin x}{\cos x} $ - Important to express all terms in sine and cosine for simplification. 3. **Start with the left-hand side (LHS):** $$\frac{1}{\sin x} - \sin x = \frac{1 - \sin^2 x}{\sin x}$$ 4. **Use the Pythagorean identity:** $$1 - \sin^2 x = \cos^2 x$$ 5. **Substitute into LHS:** $$\frac{\cos^2 x}{\sin x}$$ 6. **Rewrite the right-hand side (RHS):** $$\frac{\cos x}{\tan x} = \frac{\cos x}{\frac{\sin x}{\cos x}} = \cos x \cdot \frac{\cos x}{\sin x} = \frac{\cos^2 x}{\sin x}$$ 7. **Since LHS = RHS, the identity is proven.** --- 1. **Problem statement:** Prove the identity $ \frac{1 - \tan^2 x}{1 + \tan^2 x} = \cos^2 x - \sin^2 x $. 2. **Recall:** - $ \tan x = \frac{\sin x}{\cos x} $ - Use Pythagorean identity: $ \sin^2 x + \cos^2 x = 1 $ 3. **Rewrite LHS using $ \tan x = \frac{\sin x}{\cos x} $:** $$\frac{1 - \left(\frac{\sin x}{\cos x}\right)^2}{1 + \left(\frac{\sin x}{\cos x}\right)^2} = \frac{1 - \frac{\sin^2 x}{\cos^2 x}}{1 + \frac{\sin^2 x}{\cos^2 x}}$$ 4. **Multiply numerator and denominator by $ \cos^2 x $:** $$\frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x}$$ 5. **Use Pythagorean identity in denominator:** $$\cos^2 x + \sin^2 x = 1$$ 6. **Simplify:** $$\frac{\cos^2 x - \sin^2 x}{1} = \cos^2 x - \sin^2 x$$ 7. **Thus, LHS = RHS, identity proven.** --- 1. **Problem statement:** Prove the identity $ 2 \sin^4 x - 3 \sin^2 x + 1 = \cos^2 x (1 - 2 \sin^2 x) $. 2. **Rewrite LHS by substituting $ y = \sin^2 x $:** $$2 y^2 - 3 y + 1$$ 3. **Rewrite RHS using $ \cos^2 x = 1 - \sin^2 x = 1 - y $:** $$ (1 - y)(1 - 2 y) = 1 - 2 y - y + 2 y^2 = 1 - 3 y + 2 y^2 $$ 4. **Compare LHS and RHS:** LHS: $$2 y^2 - 3 y + 1$$ RHS: $$1 - 3 y + 2 y^2$$ 5. **Since addition is commutative, RHS = LHS, identity holds.** **Final answers:** All three identities are proven true.