Prove that :- \( \frac{10 \cos x}{1-\sin x}=\frac{1+\sin x}{\cos x} \)

We begin with the equation \[ \frac{10\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x}. \] **Step 1. Multiply Both Sides by the Denominators** Assuming \(1-\sin x \neq 0\) and \(\cos x \neq 0\), we multiply both sides by \((1-\sin x)\cos x\) to eliminate the fractions. This gives \[ 10\cos x \cdot \cos x = (1+\sin x)(1-\sin x). \] In other words, \[ 10\cos^2 x = (1+\sin x)(1-\sin x). \] **Step 2. Simplify the Right Side** Recall the difference of squares identity: \[ (1+\sin x)(1-\sin x)=1-\sin^2 x. \] Thus, the equation becomes \[ 10\cos^2 x = 1-\sin^2 x. \] **Step 3. Use the Pythagorean Identity** We know from the Pythagorean identity that \[ \sin^2 x + \cos^2 x=1, \] which can be rearranged to give \[ 1-\sin^2 x=\cos^2 x. \] Substitute this into our equation: \[ 10\cos^2 x = \cos^2 x. \] **Step 4. Solve the Equation** Subtract \(\cos^2 x\) from both sides: \[ 10\cos^2 x -\cos^2 x = 0, \] \[ 9\cos^2 x = 0. \] Divide by 9 (which is nonzero): \[ \cos^2 x = 0. \] Taking the square root on both sides, we find \[ \cos x = 0. \] **Step 5. Check for Validity in the Original Equation** The solutions to \(\cos x = 0\) are given by \[ x = \frac{\pi}{2} + k\pi,\quad k\in\mathbb{Z}. \] Now, if \(\cos x = 0\), then the denominator on the right-hand side of the original equation, \(\cos x\), becomes 0. This makes the expression \(\frac{1+\sin x}{\cos x}\) undefined. Similarly, for \(x=\frac{\pi}{2}+k\pi\), it is possible that the denominator on the left-hand side, \(1-\sin x\), may also be 0 (for example, when \(x=\frac{\pi}{2}\), \(\sin x=1\) so \(1-\sin x=0\)). Thus, even though the algebraic manipulation led us to the condition \(\cos x=0\), the values satisfying this condition are not allowed in the original equation because they make one or both denominators equal to zero. **Conclusion** The equation \[ \frac{10\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x} \] leads (under the assumption that the denominators are nonzero) to an impossible conclusion (i.e. \(10\cos^2 x=\cos^2 x\) implies \(10=1\) unless \(\cos x=0\)), and the only candidate solutions (those with \(\cos x=0\)) are not valid since they make the denominators zero. Therefore, the given equation has no valid solution.