I hope you can read this, and I hope you understand how I did 7. I probably overworked it, and it was possibly a simpler way to get it.
I cant rly do all the work rn buy if you add them together and set them equal to 180° you’ll get the answer
Answer:
48
Step-by-step explanation:
Look at the attachment
Rewrite the limand as
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = (1 - sin(<em>x</em>)) / (cos²(<em>x</em>) / sin²(<em>x</em>))
… = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / cos²(<em>x</em>)
Recall the Pythagorean identity,
sin²(<em>x</em>) + cos²(<em>x</em>) = 1
Then
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / (1 - sin²(<em>x</em>))
Factorize the denominator; it's a difference of squares, so
1 - sin²(<em>x</em>) = (1 - sin(<em>x</em>)) (1 + sin(<em>x</em>))
Cancel the common factor of 1 - sin(<em>x</em>) in the numerator and denominator:
(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = sin²(<em>x</em>) / (1 + sin(<em>x</em>))
Now the limand is continuous at <em>x</em> = <em>π</em>/2, so
