Asked by anjelica318
1. (1-sin²B)/(sin²Bcos²B)= csc²B-sec²B
--that's really the given. please help.
--that's really the given. please help.
Answers
Answered by
drwls
The left side is
cos^2 B/(sin^2 B*cos^2 B) = 1/sin^2 B
= csc^2 B
That does not equal the right side unless sec B = 0, which is not possible. It is not a valid identity.
You can prove that to yourself by picking any angle B. Let's take 30 degrees.
sin^2 30 = 1/4
cos^2 30 = 3/4
csc^2 30 = 4
sec^2 30 = 4/3
(3/4)/[3/16] = 4 is the left side
4 - 4/3 = 5/3 is the right side
cos^2 B/(sin^2 B*cos^2 B) = 1/sin^2 B
= csc^2 B
That does not equal the right side unless sec B = 0, which is not possible. It is not a valid identity.
You can prove that to yourself by picking any angle B. Let's take 30 degrees.
sin^2 30 = 1/4
cos^2 30 = 3/4
csc^2 30 = 4
sec^2 30 = 4/3
(3/4)/[3/16] = 4 is the left side
4 - 4/3 = 5/3 is the right side
Answered by
drwls
Obviously my last calculation is wrong and should be
4 - 4/3 = 8/3
In any case, the "identity" is not valid
4 - 4/3 = 8/3
In any case, the "identity" is not valid
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