1. The problem is to solve the equation $$\sin\left(A + \frac{\pi}{2}\right) = \cos\left(\frac{A}{2}\right) + 2$$ for $$0 \leq A \leq 2\pi$$. 2. Recall the identity $$\sin\left(x + \frac{\pi}{2}\right) = \cos x$$. Applying this, the equation becomes: $$\cos A = \cos\left(\frac{A}{2}\right) + 2$$ 3. Note that the range of $$\cos\theta$$ is $$[-1,1]$$, so $$\cos\left(\frac{A}{2}\right) + 2 \geq 1$$ always, but $$\cos A$$ cannot be greater than 1. Since $$\cos\left(\frac{A}{2}\right) + 2 \geq 1$$, the right side is at least 1, but the maximum value of the left side is 1. 4. For equality, $$\cos A = 1$$ and $$\cos\left(\frac{A}{2}\right) + 2 = 1$$ must hold simultaneously. 5. From $$\cos A = 1$$, we get $$A = 0, 2\pi$$ within the interval. 6. From $$\cos\left(\frac{A}{2}\right) + 2 = 1$$, we get $$\cos\left(\frac{A}{2}\right) = -1$$, which implies $$\frac{A}{2} = \pi$$ or $$A = 2\pi$$. 7. Check if these values satisfy the original equation: - At $$A=0$$: Left side $$\sin\left(0 + \frac{\pi}{2}\right) = \sin\frac{\pi}{2} = 1$$; Right side $$\cos 0 + 2 = 1 + 2 = 3$$, no equality. - At $$A=2\pi$$: Left side $$\sin\left(2\pi + \frac{\pi}{2}\right) = \sin\frac{\pi}{2} = 1$$; Right side $$\cos \pi + 2 = -1 + 2 = 1$$, equality holds. 8. Therefore, the only solution is $$A = 2\pi$$. --- 1. The problem is to show that $$\sqrt{(\csc \phi - 1)(\csc \phi + 1)} = \cot \phi$$. 2. Recall the identity $$a^2 - b^2 = (a - b)(a + b)$$. Applying this to the expression under the square root: $$\sqrt{(\csc \phi)^2 - 1^2} = \sqrt{\csc^2 \phi - 1}$$ 3. Using the Pythagorean identity: $$\csc^2 \phi = 1 + \cot^2 \phi$$ 4. Substitute into the expression: $$\sqrt{1 + \cot^2 \phi - 1} = \sqrt{\cot^2 \phi}$$ 5. Simplify the square root: $$\sqrt{\cot^2 \phi} = |\cot \phi|$$ 6. Since the domain of $$\phi$$ is not specified, and typically $$\cot \phi$$ can be positive or negative, the expression equals $$|\cot \phi|$$. 7. If $$\cot \phi$$ is positive in the domain considered, then: $$\sqrt{(\csc \phi - 1)(\csc \phi + 1)} = \cot \phi$$. Final answers: - (a) $$A = 2\pi$$ - (b) $$\sqrt{(\csc \phi - 1)(\csc \phi + 1)} = \cot \phi$$ (assuming $$\cot \phi \geq 0$$)