Question

maths topper please help to prove this...​

Answer 1

Answer:

ANSWER IS 1

Step-by-step explanation:

$y = ({ \cos(x) })^{( \frac{\pi}{2} - x)} \\ \\ log_{e}(y) = ( \frac{\pi}{2} - x)( log_{e}( \cos(x) ) \\ \\ log_{e}(y) = \frac{ ( \frac{\pi}{2} - x)}{ \frac{1}{ log_{e}( \cos(x) ) } } \: \\ \\ by \: l \: hospital \: rule \\ \\ log_{e}(y) = \frac{ - 1}{ \frac{ - 1}{ { log_{e}( \cos(x) ) }^{2} } \times - \sin(x) } \\ \\ log_{e}(y) = \sin(x) { log_{e}( \cos(x) ) }^{2} \\ \\ log_{e}(y) = ( 1)( log_{e}(( 1 ) ) \\ \\ log_{e}(y) = 0 \\ \\ therefore \: \: \: y = {e}^{0} \: \: when \: x \: tends \: to \: \frac{\pi}{2} \\ \\ y = 1.$