Originally Posted by

**oddball7465** #2

1-sin^2(2x)-sin^4(x)

=1-(sin 2x)^2-sin^4(x)

=1-(2 sin x cos x)^2-sin^4(x)

=1-4 sin^2 x cos^2 x-sin^4(x)

=1-4 sin^2 x (cos^2 x)-sin^4(x)

=1-4 sin^2 x (1- sin^2 x)-sin^4(x)

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

=1-4 sin^2 x+ 3 sin^4 x

=1-3 sin^2 x-sin^2 x + 3 sin^4 x

=(1-3 sin^2 x)-sin^2 x(1- 3 sin^2 x)

=(1-3 sin^2 x)(1 -sin^2 x)

=(1-3 sin^2 x)cos^2 x

I'm so lost on this one

#3

[cot(x)+csc(x)][cot(x)-csc(x)]

cot^2(x)-csc^2(x)

[csc^2(x)]/[sec^2(x)]-[csc^2(x)]/1

[csc^2(x)]/[sec^2(x)]-[csc^2(x)][sec^2(x)]/[sec^2(x)]

Where to now though?

#4

[3-3sin(x)][3+3sin(x)]

9[1-sin(x)][1+sin(x)]

9[sin^2(x)]

=9cos^2(x)

#5

1/[sec(x)+1]-1/[sec(x)-1]

[sec(x)-1]/[sec^2(x)-1]-[sec(x)+1]/[sec^2(x)-1]

Now what though?

#6

[cos(x)]/[1+sin(x)]+[1+sin(x)]/[cos(x)]

=[cos^2 (x)+ (1+ sin x)^2}/[1+sin(x)][cos(x)]

=[cos^2 (x)+ 1+ 2 sin x+ sin^2 x } / [1+sin(x)][cos(x)]

=[(cos^2 (x)+ sin^2 x) + 1+ 2 sin x} / [1+sin(x)][cos(x)]

=[1+ 1+ 2 sin x} / [1+sin(x)][cos(x)]

=[2+ 2 sin x} / [1+sin(x)][cos(x)]

=2[1+ sin x} / [1+sin(x)] [cos(x)]

=2/ cos(x)

= 2 sec x