A) 1
B) 0
C) \[\infty \]
D) does not exist
Correct Answer: D
Solution :
[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 .
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