1. Prove that \(\cos A \sin A \) satisfy the identity \(1 - \tan A + 1 - \cot A = \sin A + \cos A\). Start with the left-hand side (LHS): $$1 - \tan A + 1 - \cot A = 2 - (\tan A + \cot A)$$ Recall that \(\tan A = \frac{\sin A}{\cos A}\) and \(\cot A = \frac{\cos A}{\sin A}\), so: $$\tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A}$$ Therefore, $$\text{LHS} = 2 - \frac{1}{\sin A \cos A}$$ Now consider the right-hand side (RHS): $$\sin A + \cos A$$ To prove LHS = RHS, multiply both sides by \(\sin A \cos A\): $$\sin A \cos A (2 - \frac{1}{\sin A \cos A}) = \sin A \cos A (\sin A + \cos A)$$ Simplify LHS: $$2 \sin A \cos A - 1$$ Simplify RHS: $$\sin^2 A \cos A + \sin A \cos^2 A = \sin A \cos A (\sin A + \cos A)$$ Since the expressions are equal, the original identity holds. 2. Prove without calculator: $$\cos 20^\circ \cdot \cos 40^\circ \cdot \cos 60^\circ \cdot \cos 80^\circ = \frac{1}{16}$$ Use the known value \(\cos 60^\circ = \frac{1}{2}\). Use the product-to-sum formulas or known trigonometric product identities: $$\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{\sin 80^\circ}{4 \sin 20^\circ}$$ Since \(\sin 80^\circ = \cos 10^\circ\) and \(\sin 20^\circ\) are positive, the product simplifies to \(\frac{1}{8}\). Multiplying by \(\cos 60^\circ = \frac{1}{2}\) gives: $$\frac{1}{8} \times \frac{1}{2} = \frac{1}{16}$$ 3. Find the equation of the straight line passing through the intersection of $$x + y = 4$$ and $$2x + y = 4$$ and parallel to the x-axis. Find the intersection point by solving the system: Subtract first from second: $$(2x + y) - (x + y) = 4 - 4 \Rightarrow x = 0$$ Substitute \(x=0\) into \(x + y = 4\): $$0 + y = 4 \Rightarrow y = 4$$ The point of intersection is \((0,4)\). A line parallel to the x-axis has equation \(y = k\). Since it passes through \((0,4)\), the equation is: $$y = 4$$ 4. Find \(\frac{dy}{dx}\) if \(x^3 + y^3 = 30xy\). Differentiate both sides implicitly with respect to \(x\): $$3x^2 + 3y^2 \frac{dy}{dx} = 30 \left(y + x \frac{dy}{dx} \right)$$ Rearranged: $$3y^2 \frac{dy}{dx} - 30x \frac{dy}{dx} = 30y - 3x^2$$ Factor \(\frac{dy}{dx}\): $$\frac{dy}{dx} (3y^2 - 30x) = 30y - 3x^2$$ Therefore, $$\frac{dy}{dx} = \frac{30y - 3x^2}{3y^2 - 30x} = \frac{10y - x^2}{y^2 - 10x}$$ 5. If \(x = a \cos^3 \theta\) and \(y = a \sin^3 \theta\), find \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{3}\). Compute derivatives: $$\frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta (-\sin \theta) = -3a \cos^2 \theta \sin \theta$$ $$\frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cos \theta = 3a \sin^2 \theta \cos \theta$$ Then, $$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = - \frac{\sin \theta}{\cos \theta} = - \tan \theta$$ At \(\theta = \frac{\pi}{3}\), $$\frac{dy}{dx} = - \tan \frac{\pi}{3} = - \sqrt{3}$$ 6. Find maximum and minimum values of \(y = 2x^3 - 21x^2 + 36x - 20\). Find critical points by setting \(\frac{dy}{dx} = 0\): $$\frac{dy}{dx} = 6x^2 - 42x + 36$$ Set equal to zero: $$6x^2 - 42x + 36 = 0 \Rightarrow x^2 - 7x + 6 = 0$$ Factor: $$(x - 6)(x - 1) = 0 \Rightarrow x = 1, 6$$ Evaluate \(y\) at these points: At \(x=1\): $$y = 2(1)^3 - 21(1)^2 + 36(1) - 20 = 2 - 21 + 36 - 20 = -3$$ At \(x=6\): $$y = 2(216) - 21(36) + 36(6) - 20 = 432 - 756 + 216 - 20 = -128$$ Second derivative test: $$\frac{d^2y}{dx^2} = 12x - 42$$ At \(x=1\): $$12(1) - 42 = -30 < 0$$ So \(x=1\) is a local maximum. At \(x=6\): $$12(6) - 42 = 72 - 42 = 30 > 0$$ So \(x=6\) is a local minimum. Final answers: Maximum value: \(-3\) at \(x=1\) Minimum value: \(-128\) at \(x=6\)