Question

Verifying Trigonometric Identies help.

Answer 1

\[\frac{ 1 }{sinx-1 }-\frac{ 1 }{ sinx+1 }=-2\sec^2x\]

Answer 2

\[\frac{ sinx+1 }{\sin^2x-1}-\frac{ sinx-1 }{\sin^2x-1 }=\frac{ 2 }{ \sin^2x-1}\]

Answer 3

\[(a + b) (a - b) = (a^2 - b^2)\]

Answer 4

that's what I got so far

Answer 5

u mean
\[\frac{ \sin x + 1 - (\sin x -1) }{ \sin^2 x - 1 }\]

Answer 6

it wouldn't just be \[\frac{ 2 }{\sin^2x-1 }\]

Answer 7

so it would be
\[\frac{ 2 }{ \sin^2 x - 1 }\]

and
cos^2 x + sin^2 x = 1
proceed

Answer 8

\[\sin^2 - 1 = - \cos^2 x \]

Answer 9

so \[\frac{ 2}{ \cos^2x }\]

Answer 10

its
\[\frac{ -2 }{ \cos^2 x }\]

Answer 11

ah yeah I see that now

Answer 12

and
\[\frac{ 1 }{ \cos x } = \sec x\]

Answer 13

I know sec x= 1/cos x

Answer 14

it still keeps the ^2 ?

Answer 15

\[\frac{ 1 }{ \cos^2 x } = \sec^2 x\]

Answer 16

thank you I got it now

Answer 17

u welcome :))