SOLUTION: please help me to answer this problem: derive the identity: cot^2 A + 1 = csc ^2 A .

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Question 86952: please help me to answer this problem: derive the identity: cot^2 A + 1 = csc ^2 A .
Found 2 solutions by Edwin McCravy, jim_thompson5910:
Answer by Edwin McCravy(3605) About Me  (Show Source):
You can put this solution on YOUR website!
please help me to answer this problem: derive the identity:
cot%5E2A+%2B+1+=+csc%5E2A .
Draw a standard labeled right triangle:

drawing%28200%2C200%2C-1%2C10%2C-1%2C15%2C+triangle%280%2C0%2C8%2C0%2C8%2C15%29%2C%0D%0Alocate%288.5%2C7.5%2C%27a%27%29%2C+locate%282.7%2C7.5%2C%27c%27%29%2C+locate%28-.5%2C0%2C%27A%27%29%2C%0D%0Alocate%288.5%2C0%2C%27C%27%29%2C+locate%288.5%2C15%2C%27B%27%29%2C+locate%284%2C-.5%2C%27b%27%29%2C%0D%0Arectangle%287.5%2C0%2C8%2C.5%29%0D%0A%0D%0A+%29
By the Pythagorean theorem:
a%5E2 + b%5E2 = c%5E2
Divide every term by a%5E2
a%5E2%2Fa%5E2 + b%5E2%2Fa%5E2 = c%5E2%2Fa%5E2
The first term, a%5E2%2Fa%5E2 is just 1.
1 + b%5E2%2Fa%5E2 = c%5E2%2Fa%5E2
Now we can write b%5E2%2Fa%5E2 as %28b%2Fa%29%5E2, and
we can write c%5E2%2Fa%5E2 as %28c%2Fa%29%5E2,
1 + %28b%2Fa%29%5E2 = %28c%2Fa%29%5E2
Now b%2Fa is %28adjacent%29%2F%28opposite%29 which is cot%28A%29, and
c%2Fa is %28hypotenuse%29%2F%28opposite%29 which is csc%28A%29,
so, upon substituting these, we have
1+%2B+cot%5E2A+=+csc%5E2A
or reversing the terms on the left:
cot%5E2A+%2B+1+=+csc%5E2A
Edwin

Answer by jim_thompson5910(14873) About Me  (Show Source): 

You can put this solution on YOUR website!
 Start with the given expression





 Replace  with 





 Replace "1" with  







 Combine the fractions





 Use the identity  to replace  with 1









 Use the identity   to replace  with 





So now our identity is proven