Express this trigonometric identity in simplest form?

2018-01-04 6:11 am
(csc^2 (x)-1)/(csc(x)-1)

回答 (12)

2018-01-04 10:06 am
(csc^2x - 1)/(cscx - 1) =
(cscx + 1)(cscx - 1)/(cscx - 1) =
cscx + 1
2018-01-05 12:15 am
(csc^2 (x) - 1)/(csc (x) - 1)
= csc (x) + 1
2018-01-04 10:29 am
It's not an identity.
2018-01-04 8:24 am
csc(x) + 1, provided that sin(x) =/= 1. Simply factor the numerator as a difference of squares, then cancel out.
2018-01-04 8:10 am
[csc^2(x)-1]/[csc(x) - 1] = csc(x) + 1]*[ csc(x) - 1]/ [csc(x) - 1],
= csc(x) + 1,
or, = [1 + sin(x)]/sin(x) ...........(Ans.)
2018-01-04 6:27 am
This is a wrongly worded question. It quotes an expression not an identity. An identity has an equals sign in it.
The question should read "simplify the expression...."
2018-01-04 6:24 am
you have an " [ a² - b² ] / [ a - b ] "...use it
2018-01-04 6:24 am
Rule
a^2-b^2=(a-b)(a+b)
2018-01-04 6:23 am
(csc^2 (x) - 1)/(csc(x) - 1)
= ((cos (x) + 1)(cos (x) - 1))/(csc(x) - 1)
= csc(x) + 1
2018-01-04 6:21 am
[ csc x - 1 ] [ csc x + 1 ]
-------------------------------- = csc x + 1
csc x - 1


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