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प्रश्न
\[\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\]
उत्तर
\[\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\]
\[ = \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin^2 x}{1 - \sin x} \binom{ \because \cos^2 x = 1 - \sin^2 x }{ a^2 - b^2 = \left( a - b \right) \left( a + b \right)}\]
\[ = \lim_{x \to \frac{\pi}{2}} \frac{\left( 1 - \sin x \right) \left( 1 + \sin x \right)}{\left( 1 - \sin x \right)}\]
\[ = \lim_{x \to \frac{\pi}{2}} \left( 1 + \sin x \right)\]
\[ \Rightarrow 1 + 1 = 2\]
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