Question

Trig identities

Answer 1

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

Answer 2

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

Answer 3

You're going to have to add the ones on the left before you see it.

Answer 4

what do you mean ? :S

Answer 5

\[\Large
\frac{a}{b} - \frac{c}{d} = \frac{ad-cb}{bd}
\]

Answer 6

for the right side i get this : \[\huge \frac{ (sinx-1)-(sinx+1) }{ 1-\sin^2x }\]

Answer 7

dont make your life hard

combine the left hand side into 1 fraction first

Answer 8

then multiply the denominators to remove the fraction

Answer 9

then it would be extremely easy to solve

Answer 10

ohhh

Answer 11

when u make the LHS into one fraction, u will get something like
\[\frac{ a }{ b }=\frac{ c }{ d }\]
\[ad = cb\]