Question

Prove this trig identity

Answer 1

\[\huge \cos2x=\frac{ 1-\tan^2x }{ 1+\tan^2x }\]

Answer 2

\[\cos(2x)=\cos^2(x)-\sin^2(x)=\cos^2(x)(1-\tan^2(x)) =\frac{1-\tan^2(x)}{\sec^2(x)}\]
\[=\frac{1-\tan^2(x)}{1+\tan^2(x)}\]

Answer 3

how does \[\huge \frac{ 1-\tan^2x }{ \sec^2x } = \frac{ 1-\tan^2x }{ 1+\tan^2x}\]

Answer 4

@malevolence19

Answer 5

Because sec^2(x) = 1 + tan^2(x)

Answer 6

It's one of the identities, you can see it here: http://www.purplemath.com/modules/idents.htm

Answer 7

or prove it like so:

\[\sec ^{2}x=\frac{ 1 }{ \cos ^{2} x}=\frac{ \sin ^{2}x+\cos ^{2}x }{ \cos ^{2}x }=\frac{ \sin ^{2}x }{ \cos ^{2}x }+1=\tan ^{2}x+1\]

Answer 8

thanks :)

Answer 9

Sorry, I was gone D: Glad you got it!