Question
dy/dx[tan(y)=x=y] in terms of y
the answer my teacher got was cot^2(y) but i get is 1/[sec^2(y)-1]
either i made a mistake, my teacher made a mistake or we're both right and i just need to simplify.
please tell me which of the following is the case and how to get to the correct answer.
the answer my teacher got was cot^2(y) but i get is 1/[sec^2(y)-1]
either i made a mistake, my teacher made a mistake or we're both right and i just need to simplify.
please tell me which of the following is the case and how to get to the correct answer.
Answers
let's see if cot^2 y = 1/(sec^2 y - 1 )
RS = 1/(1/cos^2y - 1)
= 1/((1 - cos^2y)/cos^2y)
= 1/((sin^2y/cos^2))
= cos^2y/sin^2y
= cot^2 y
so you are both right!
RS = 1/(1/cos^2y - 1)
= 1/((1 - cos^2y)/cos^2y)
= 1/((sin^2y/cos^2))
= cos^2y/sin^2y
= cot^2 y
so you are both right!
ok, thank you. is it preferable to change it like that? cause i honestly have no idea why it would matter.
oh and just to be clear you got this using trig identities right?
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