I'm in grade 11, and doing a summative project. The question asks to prove that:
sin^2(x)+2cosx-1/sin^2(x)+3cosx-3
=cos^2(x)+cosx/-sin^2(x)
I need to do this using the quotient, pythagorean, and reciprocal identities. I have changed sin^2(x) to 1+cos^2(x), and done the same to cos^2(x). Where can I go from here?
Any help is appreciated. Thanks :)How do I solve this trigonometric identity?
I assume that brackets go like this:
{sin^2(x)+2cosx-1}/{sin^2(x)+3cosx-3}
sin^2(x) = 1- cos^2(x); - not +
{sin^2(x)+2cosx-1}/{sin^2(x)+3cosx-3}
= {1- cos^2(x)+2cosx-1}/{1- cos^2(x)+3cosx-3}
= {- cos^2(x)+2cosx}/{- cos^2(x)+3cosx-2}
= { cos^2(x)-2cosx}/{ cos^2(x)-3cosx+2}
= cosx(cosx-2)/{(cosx-1)(cosx-2)}
=cosx/(cosx-1) : multiply top and bottom by cosx+1
=(cosx+1)cosx/(cosx-1)(cosx+1)
={cos^2(x)+cosx}/{cos^2(x)-1}
={cos^2(x)+cosx}/{-sin^2(x)}
