(tan(<em>x</em>) + cot(<em>x</em>)) / (tan(<em>x</em>) - cot(<em>x</em>)) = (tan²(<em>x</em>) + 1) / (tan²(<em>x</em>) - 1)
… = (sin²(<em>x</em>) + cos²(<em>x</em>)) / (sin²(<em>x</em>) - cos²(<em>x</em>))
… = -1/cos(2<em>x</em>)
Then as <em>x</em> approaches <em>π</em>/2, the limit is -1/cos(2•<em>π</em>/2) = -sec(<em>π</em>) = 1.
Answer:
Step-by-step explanation:
Given that the random variable X is normally distributed, with
mean = 50 and standard deviation = 7.
Then we have z= 
Using this and normal table we find that
a) 
b) When z=0.02
we get

c) 90th percentile z value =1.645
90th percentile of X 
The Answer Is 57.92 But the easiest way to do it is Rounding it