Question

how would i prove the identity (sin^2x - 6sinx+9)/sin^2x-9 = (sinx-3)/sinx+3

Answer 1

look at factorising the left hand side

(sin^2(x) - 6sin(x) + 9 = (sin(x) -3)^2

sin^(x) - 9 = (sin(x)-3)(sin(x) + 3)

take sin(x) -3 out as a common factor

\[\frac{(\sin(x)-3)^2}{(\sin(x)-3)(\sin(x)+3)} = \frac{\sin(x) -3}{\sin(x) +3}\]

Answer 2

= (sin x - 3)^2
-----------
(sinx -3 )(sinx + 3)

cancelling out sinx - 3 gives the desired result

Answer 3

@campbell_st @joeywhite thank you :)