Question

Verify. ( I know my identities)
cscx-cotx=1/cscx+cotx

Answer 1

subtract

Answer 2

HI!!

Answer 3

you can maybe work with the right and left
the right is
\[\frac{1}{\frac{1}{b}+\frac{a}{b}}\] the left is
\[\frac{1}{b}-\frac{a}{b}\]

Answer 4

most of it is probably algebra

Answer 5

\[\frac{1}{b}-\frac{a}{b}=\frac{1-a}{b}\]

Answer 6

\[\frac{1}{\frac{1}{b}+\frac{a}{b}}\]\[=\frac{1}{\frac{1+a}{b}}\]\[=\frac{b}{1+a}\]

Answer 7

so now we are at
\[\frac{1-\cos(x)}{\sin(x)}\] on the the right, and
\[\frac{\sin(x)}{1+\cos(x)}\] on the left

Answer 8

that is a well known identity, but you can prove that without much problem if it is not

Answer 9

cscx-cotx = 1/(cscx+cotx)
--------------
1 cscx - cot x
----- * ------------
csc x + cot x csc x - cot x

csc x - cot x
------------
csc^ 2x - cot^2x

Recall the Pythagorean Identity: cscx - cot x
------------
(1 + cot^2x) - cot^2x
1 + cot^2x = csc^2x
cscx - cot x
--------------
1

csc x - cot x

Answer 10

@JMAC1204