Question

Verify 2cos2x/sin2x=cotx-tanx

Answer 1

is it cos^2x or cos 2x.

Answer 2

cos2x. no exponent just number

Answer 3

ok.

so cos 2x = 2 cos^2x - 1 which is 2 cos^2x - (sin^2x+cos^2x)

Hence cos 2x = cos^2x - sin^2x.

also, sin 2x = 2 sin x cos x.

So we can simplify the fraction as -

\[\frac{ 2(\cos^2x-\sin^2x) }{ 2 \sin x \cos x } => \frac{ (\cos^2x-\sin^2x }{ \sin x \cos x }\]

Answer 4

ok till here?

Answer 5

yes! thank you!

Answer 6

Now you can separate the fractions -

\[\frac{ \cos^2x }{ \sin x \cos x } - \frac{ \sin^2x }{ \sin x \cos x }\]

Answer 7

cancel the cos in LH and sin in RH. you'll get your result. :)

Answer 8

thank you so much