# TRIG IDENTITY! math help please!?

please prove

[cos4x•tan2x - sin4x] / [cos4x•cot2x + sin4x] = -tan^2(2x)

please show as much work as possible :P thanks so much for your time! I'll select as best answer as soon as yahoo allows me!

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### 1 Answer

- 7 years agoFavorite Answer
hey that was easy than i thought

i thought it will be difficult :)

but it's not :)

finally i did it XD

solution-

LHS:

let do it part by part :)

numerator-

[cos4x•tan2x - sin4x]

(cos4x * sin2x/cos2x - sin4x)

(cos4x * sin2x - sin4x * cos2x)/cos2x

use identity

sin(α – β) = sin(α)cos(β) – cos(α)sin(β)

sin(2x-4x)/cos2x

-sin 2x / cos 2x

-tan 2x

denominator:

[cos4x•cot2x + sin4x]

cos4x * cos 2x/sin2x + sin4x

cos(α – β) = cos(α)cos(β) + sin(α)sin(β)

(cos 4x * cos2x + sin 4x * sin 2x)/sin 2x

cos(4x-2x)/sin2x

cos2x/sin2x

=cot2x

-tan 2x/cot2x

-tan2x/(1/tan2x)

-tan^2x <=== answer :)

Source(s): grade 12 now :)- Login to reply the answers