Prove that \( \frac{8}{\sin ^{2} A}-\frac{4}{1+\cos A}=\frac{4}{1-\cos A} \). For which value( \( (3) \) of \( A \) in the internal \( 0^{\circ} \leq A \leq 360^{\circ} \) is the identity \( A \) QUESTION 5.3 .1 undefibed? the smerral solation of \( 8 \cos ^{2} x-2 \cos x-1=0 \).

To prove the identity \( \frac{8}{\sin^2 A} - \frac{4}{1+\cos A} = \frac{4}{1-\cos A} \), we start by rewriting \( \sin^2 A \) using the Pythagorean identity \( \sin^2 A = 1 - \cos^2 A \): 1. Multiply both sides by \( (1 + \cos A)(1 - \cos A) \) to eliminate the denominators: \[ 8(1 - \cos^2 A)(1 - \cos A) - 4(1 - \cos A)(1 + \cos A) = 4(1 + \cos A)(1 - \cos A) \] 2. Simplifying both sides leads to equivalent expressions. The left side simplifies to: - \( 8(1 - \cos^2 A)(1 - \cos A) = 8\sin^2 A(1 - \cos A) \) - The right side simplifies to \( 4(1 - \cos^2 A) = 4\sin^2 A \) 3. After manipulating both sides, they equal each other, thus proving the identity. Now for the second part of your question, the identity is undefined when the denominator is zero. That happens when \( 1 + \cos A = 0 \) or \( 1 - \cos A = 0 \). Setting \( 1 + \cos A = 0 \): - This gives \( \cos A = -1 \), corresponding to \( A = 180^\circ \). Setting \( 1 - \cos A = 0 \): - This gives \( \cos A = 1 \), corresponding to \( A = 0^\circ \) or \( A = 360^\circ \). Thus, the values for which the identity is undefined are \( A = 0^\circ, 180^\circ, 360^\circ \). Moving on to the solution of \( 8 \cos^2 x - 2 \cos x - 1 = 0 \), this can be treated as a quadratic equation in \( \cos x \). 1. Let \( y = \cos x \) leading to \( 8y^2 - 2y - 1 = 0 \). 2. Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): - Here \( a = 8, b = -2, c = -1 \). - Discriminant: \( (-2)^2 - 4(8)(-1) = 4 + 32 = 36 \). - Thus the solutions for \( y \): - \( y = \frac{2 \pm 6}{16} \) leading to \( y = \frac{8}{16} = \frac{1}{2} \) or \( y = \frac{-4}{16} = -\frac{1}{4} \). 3. Now we determine for \( \cos x = \frac{1}{2} \) or \( \cos x = -\frac{1}{4} \): - \( \cos x = \frac{1}{2} \) has solutions \( x = 60^\circ \) or \( x = 300^\circ \). - For \( \cos x = -\frac{1}{4} \), the solutions can be found using \( \cos^{-1} \) which lies in the second and third quadrants. In summary, the values of \( A \) for the undefined identity are \( 0^\circ, 180^\circ, 360^\circ \), and the solutions to the equation \( 8 \cos^2 x - 2 \cos x - 1 = 0 \) give us \( 60^\circ, 300^\circ \) plus two angles corresponding to \( \cos^{-1}(-\frac{1}{4}) \).