If \(\mathrm{y}=\mathrm{x} \tan (\mathrm{x} / 2)\) and \(\mathrm{A}(\mathrm{dy} / \mathrm{dx})-\mathrm{B}=\mathrm{x}\) then \((\mathrm{B} / \mathrm{A})=\) (a) \(\cot (\mathrm{x} / 2)\) (b) \(\tan (\mathrm{x} / 2)\) (c) \(\tan \mathrm{x}\) (d) \(\cot x\)

To find the derivative of y = x * tan(x/2), we need to use the product rule, which states that the derivative of f(x) * g(x) is f'(x) * g(x) + f(x) * g'(x). In this case, \(f(x) = x\) and \(g(x) = \tan(x/2)\). First, let's differentiate f(x) and g(x): \(f'(x) = \frac{d}{dx}(x) = 1\) \(g'(x) = \frac{d}{dx}(\tan(x/2))\) (we'll use the chain rule here) The chain rule states that the derivative of a composite function h(g(x)) is h'(g(x)) * g'(x). In this case, \(h(u) = \tan(u)\) and \(u(x) = x/2\). To apply the chain rule, we first need to find the derivatives of h(u) and u(x): \(h'(u) = \frac{d}{du}(\tan(u)) = \sec^2(u)\) \(u'(x) = \frac{d}{dx}(x/2) = 1/2\) Now we can apply the chain rule and find the derivative of g(x): \(g'(x) = h'(x/2) * u'(x) = \sec^2(x/2) * (1/2)\) Now that we have the derivatives of f(x) and g(x), we can use the product rule to find dy/dx: \(dy/dx = f'(x) * g(x) + f(x) * g'(x) = 1 * \tan(x/2) + x * \frac{1}{2} \sec^2(x/2)\)

The given equation is A(dy/dx)-B = x. We can substitute our expression for dy/dx: \(A(1 * \tan(x/2) + x * \frac{1}{2} \sec^2(x/2)) - B = x\)

We are tasked with finding B/A. First, we need to isolate B on one side of the equation. To do that, let's factor A out of the left-hand side of the equation: \(A (\tan(x/2) + \frac{x}{2} \sec^2(x/2)) - B = x\) Now, we can rearrange the equation to isolate B: \(B = A (\tan(x/2) + \frac{x}{2} \sec^2(x/2)) - x\) Now, we can easily find B/A: \(\frac{B}{A} = \frac{A (\tan(x/2) + \frac{x}{2} \sec^2(x/2)) - x}{A}\) When we cancel A in the numerator and denominator, we get: \(\frac{B}{A} = \tan(x/2) + \frac{x}{2} \sec^2(x/2) - \frac{x}{A}\) This expression has a term \(\frac{x}{A}\) left. The only way for this expression to be equal to one of the given options is when \(\frac{x}{A} = 0\). That means B/A can be simplified to: \(\frac{B}{A} = \tan(x/2) + \frac{x}{2} \sec^2(x/2)\) Since there is no option that exactly matches this expression, there might be a mistake in the given exercise or options.