Prove that? # (cos^4 x - sin^4 x)/(1-tan x) -= cos^2 x +sin x cos x #
2 Answers
Please see proof below.
Explanation:
We seek to prove that the following is an indetiuty:
# (cos^4 x - sin^4 x)/(1-tan x) -= cos^2 x +sin x cos x #
Consider the LHS::
# LHS -= (cos^4 x - sin^4 x)/(1-tan x) #
# \ \ \ \ \ \ \ \ = ((cos^2 x)^2 - (sin^2 x)^2)/(1-tan x) #
This is the difference of two squares, so we can use the identity:
# A^2 - B^2 -= (A+B)(A-B) #
Thus:
# LHS = ((cos^2 x+sin^2 x)(cos^2 x-sin^2 x))/(1-tan x) #
Now, And again, we have the difference of two squares, so we can factor further, and also we use the trigonometric identity:
# sin^2x+cos^2x -= 1 #
Thus:
# LHS = ((1)(cosx+sinx)(cos x-sin x))/(1-tan x) #
# \ \ \ \ \ \ \ \ = ((cosx+sinx)(cos x-sin x))/(1-sinx/cosx) #
# \ \ \ \ \ \ \ \ = ((cosx+sinx)(cos x-sin x))/((cosx-sinx)/cosx) #
# \ \ \ \ \ \ \ \ = ((cosx+sinx)(cos x-sin x)cosx)/((cosx-sinx)) #
# \ \ \ \ \ \ \ \ = (cosx+sinx)cosx #
# \ \ \ \ \ \ \ \ = cos^2x+sinxcosx #
# \ \ \ \ \ \ \ \ = RHS \ \ \ \ # QED