Question

If $f\left(x\right) = \frac{1-cos\,x}{1-sin\,x}$, then $f' \left(\frac{\pi}{2}\right)$ is equal to

• A. $1$
• B. $0$
• C. $\infty$
• D. does not exist

Answer 1

Answer: (D)

$f\left(x\right) = \frac{\left(1-sin\,x\right)\left(sin\,x\right)-\left(1-cos\, x\right)\left(-cos\,x\right)}{\left(1-sin\,x\right)^{2}}$
$= \frac{sin\,x - sin^{2}\,x+cos\, x-cos^{2}\,x}{\left(1-sin\,x\right)^{2}} = \frac{sin\,x+cos\,x - 1}{\left(1-sin\,x\right)^{2}}$
$\therefore \quad f'\left(\frac{\pi}{2}\right) = \frac{1+0-1}{\left(1-1\right)^{2}} = \frac{0}{0}$
Therefore, $f'\left(\frac{\pi }{2}\right)$ does not exist.