Question

(tan^3x-cot^3x)/(tan^2x+csc^2x)=tanx-cotx I need to prove and oonly work with one side to do so. I am working with the left side.

Answer 1

Or if you can figure out form the right side that would help too

Answer 2

\[\frac{\tan^3x-\cot^3x}{\tan^2x+\csc^2x}=\tan x-\cot x\]
Factor the difference of cubes in the numerator:
\[a^3-b^3=(a-b)\left(a^2+ab+b^2\right)\]
Using an identity or two, you can rewrite the numerator so that the second factor cancels out with the denominator.

Answer 3

\[\tan^3x -\cot^3x = (tanx-cotx)(\tan^2x+tanx*cotx+\cot^2x)\]
as we have cotx = 1/tanx then tanx.cotx =1
\[(tanx-cotx)(\tan^2x+\cot^2x+1)\]
and we have an identity
\[cosec^2x-\cot^2x =1\]
using this
\[cosec^2x = 1+\cot^2x\]
substituting this resulf of 1 +cot^2x in previous one we get
\[(tanx-cotx)(\tan^2x+cosec^2x)\]
now
\[\frac{ (tanx-cotx)(\tan^2x+cosec^2x) }{ \tan^2x+cosec^2x }\]
will give you tanx - cotx

Answer 4

Thank you so much!!

Answer 5

Its my pleasure ( and pain too ! :( )

Answer 6

If u have any kind of doubt in trigonometry, u can count on me!