• # question_answer If $\mathbf{f}\left( \mathbf{x} \right)=\frac{\mathbf{1}-\mathbf{cosx}}{\mathbf{1}-\mathbf{sinx}}$ then $f'\left( \frac{\pi }{2} \right)$is equal to A)  1                                 B)         0               C)         $\infty$                                   D)         does not exist

[d]$f(x)=\frac{1-cosx}{1-\sin x}$ $\left( 1-sinx \right).\frac{d}{dx}\left( 1-cosx \right)$ $\therefore f'(x)=\frac{-\left( 1-cosx \right)\times \frac{d}{dx}\left( 1-sinx \right)}{{{\left( 1-sinx \right)}^{2}}}$ $=\frac{\left( 1-sinx \right).sinx-\left( 1-cosx \right)\left( -cosx \right)}{{{\left( 1-sinx \right)}^{2}}}$ $\frac{sinx-si{{n}^{2}}x+cosx-co{{s}^{2}}x}{{{\left( 1-sinx \right)}^{2}}}=\frac{\left( sinx+cosx \right)-\left( si{{n}^{2}}x+co{{s}^{2}}x \right)}{{{\left( 1-sinx \right)}^{2}}}$$f'\left( \frac{\pi }{2} \right)=\frac{\sin \frac{\pi }{2}+\cos \frac{\pi }{2}-1}{{{\left( 1-\sin \frac{\pi }{2} \right)}^{2}}}=\frac{0}{0}$ It doesn't exist. Hence, the correct option is .