Question

Trig identity.. I don't understand this one at all

Answer 1

how do i make the left side into one big fraction

Answer 2

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

Answer 3

typed it wrong wooops

Answer 4

get common denominators on the left side:
\(\huge \frac{ 1 }{ sinx+1 }\cdot (\color {red}{\frac{ sinx-1 }{ sinx-1 }}) - \frac{ 1 }{ sinx-1 }\cdot( \color {red}{\frac{ sinx+1 }{ sinx+1 }}) = \frac{ 2 }{ 1-\sin^2x} \)

Answer 5

so the left side in simplified form will be:
\(\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1} \)
can you simplify this?

Answer 6

no that's where i get stuck
how would i subtract the numerator

Answer 7

just use distributive property to get rid of the parenthesis...

Answer 8

@dpaInc i get this \[\huge \frac{ \sin^2x+sinx-sinx-1 }{ sinx^2-1 }\]

Answer 9

how did you get that? all you're doing is subtracting....:
\(\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1} \)
=\(\huge \frac{1\cdot (sinx-1)-1\cdot (sinx+1)}{sin^2x-1} \)
=\(\huge \frac{sinx-1-sinx-1}{sin^2x-1} \)
=\(\huge \frac{-2}{sin^2x-1} \)
=\(\huge \frac{2}{1-sin^2x} \)

Answer 10

sinx-1-sinx-1 = -2

is that a trig rule or what ? :S