Question

PLEASE HELP
Verify the identity
(cotx+tanx)cscxsecx=sec^2x+csc^2x

Answer 1

ick

Answer 2

my bet is that this has very little to do with trig and a lot to do with algebra
best bet is to rewrite everything in terms of sine and cosine, do the algebra and see if it works out
want to try it?

Answer 3

oh yes, it is algebra for sure

Answer 4

it will be a lot easier to write if we replace \(\cos(x)\) by \(a\) and \(\sin(x)\) by \(b\)
then we can rewrite the left hand side as
\[\frac{\frac{a}{b}+\frac{b}{a}}{\frac{1}{ab}}\]
is that clear so far?

Answer 5

let me know if it is or is not, then we can continue
only one algebra step to finish this

Answer 6

im sorry im lost. i dont get anything that deals with this.

Answer 7

ok we can go slow
do you know what \(\cot(x)\) is in terms of \(\sin(x)\) and \(\cos(x)\)? in other words, the basic definition of cotangent?

Answer 8

"no" is a perfectly good answer, i can tell you if you like

Answer 9

start from the left hand side
(cot + tan) csc sec = cot csc sec + tan csc sec ( I just open the bracket)
the first term is cot csc sec = \(\dfrac {cos}{sin}*csc *\dfrac{1}{cos}= \dfrac{1}{sin}*csc =csc^2\)
the second term is tan csc sec = \(\dfrac{sin}{cos}*\dfrac{1}{sin}*sec=\dfrac{1}{cos}*sec = sec^2\)
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add 2 terms, you get the right hand side.