Find the approximate increment of y = (1+ cos x)/(1- cos

Find the approximate increment of $y = \dfrac{1+\cos x}{1-\cos x}$ when $x$ changes from $\dfrac{\pi}{3}$ to $\dfrac{\pi}{3}+\dfrac{1}{100}$.

Step 1: Differentiate by the quotient rule

With $u = 1+\cos x,\ u' = -\sin x$ and $v = 1-\cos x,\ v' = \sin x$: $y' = \frac{-\sin x\bigl(1-\cos x\bigr)-\bigl(1+\cos x\bigr)\sin x}{\bigl(1-\cos x\bigr)^{2}}$.

Step 2: Factor the numerator

Both terms carry $-\sin x$; the two brackets then add to $2$, and the cosines cancel: \[-\sin x\Bigl[\bigl(1-\cos x\bigr)+\bigl(1+\cos x\bigr)\Bigr] = -2\sin x, \qquad\text{so}\qquad y' = \frac{-2\sin x}{\bigl(1-\cos x\bigr)^{2}}.\]

Step 3: Evaluate at $x_{0} = \dfrac{\pi}{3}$

\[ \sin\frac{\pi}{3} = \frac{\sqrt3}{2}, \qquad 1-\cos\frac{\pi}{3} = 1-\frac12 = \frac12, \qquad \left(\frac12\right)^{2} = \frac14. \]

\[ y'\Bigl|_{\pi/3} = \frac{-2\cdot\dfrac{\sqrt3}{2}}{\dfrac14} = -4\sqrt3 \approx -6.9282. \]

Step 4: Apply the differential

With $\Delta x = \dfrac{1}{100}$: \[\Delta y \approx dy = -4\sqrt3\cdot\frac{1}{100} = -\frac{4\sqrt3}{100} = -\frac{\sqrt3}{25}.\]

Step 5: Verification

\[y\left(\frac{\pi}{3}\right) = \frac{1+\tfrac12}{1-\tfrac12} = \frac{3/2}{1/2} = 3\]

\[y\left(\frac{\pi}{3}+0.01\right) \approx 2.9317054, \qquad \Delta y_{\text{exact}} \approx -0.0682946\]

\[\text{Error of the estimate: } 0.00099, \text{ about } 1.4\% \text{ of } \Delta y \;\;\checkmark\]