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प्रश्न
Evaluate `lim_(x -> pi/2) (secx - tanx)`
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उत्तर
Put `y = pi/2 - x`.
Then y → 0 as x → `pi/2`.
Thereofre `lim_(x -> pi/2) (secx - tanx)`
= `lim_(y -> 0) [sec(pi/2 - y) - tan(pi/2 - y)]`
= `lim_(y -> 0) 1/siny = cosy/siny`
= `lim_(y -> 0) (1 - cosy)/siny`
= `lim_(y -> 0) (2sin^2 y/2)/(2sin y/2 cos y/2)`
Since, `sin^2 y/2 = (1 - cosy)/2`
sin y = `2sin y/2 cos y/2`
= `lim_(y/2 _> 0) tan y/2` = 0
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