Question

Is this an identity ? Thank you :)

4sin^2x -1
---------- = 2 sin x-1
2sin^2 x +1

Answer 1

no
\[\frac{ a ^{2}-b ^{2} }{ a+b }=\frac{ \left( a-b \right)\left( a+b \right) }{a+b }=a-b\]

Answer 2

still not clear for me

Answer 3

Identities are those expressions which are always true.... Have a look at the left side of the equation you have.....See if you can substitute it in the expression that surijathyer gave and if you get the right hand side.....If not it is not an identity

Answer 4

I see, therefore its not.
thanks :)

Answer 5

You are welcome

Answer 6

\[\frac{ \left( 2\sin ^{2}x \right)^{2}-1^{2} }{2\sin ^{2}x+1 }=\frac{ \left( 2\sin ^{2}x+1 \right)\left( 2 \sin ^{2}x-1 \right)}{ 2 \sin ^{2}x+1 }\]
\[2 \sin ^{2}x-1\]