Solve and prove the identity
(tan x+ cot (-x))/ (tan x - cot(-x))= 1-2cos^2(x)
recall that cot -x = -cotx = -1/tanx
(sin/cos + cos-/sin-)/[sin/cos-cos-/sin-]
sin -x = -sin x
cos -x = cos x
(sin/cos - cos/sin)/[sin/cos+cos/sin]
multiply top and bottom by sin cos
(sin^2 - cos^2)/(sin^2 +cos^2)
but sin^2+cos^2 = 1
sin^2-cos^2
but sin^2=1-cos^2
1 - cos^2 -cos^2
1 - 2 cos^2
well, that is what is on the right :)
To solve and prove the given identity, let's start by simplifying the left-hand side (LHS) of the equation:
LHS = (tan x + cot (-x))/ (tan x - cot (-x))
Recall that cot (-x) = 1/tan (-x), and tan (-x) = -tan x, which means that cot (-x) = -1/tan x.
Now we can substitute these values into the expression:
LHS = (tan x - (1/tan x))/ (tan x - (-1/tan x))
= (tan x + 1/tan x)/ (tan x + 1/tan x)
= 1
Thus, the LHS simplifies to 1.
Next, let's simplify the right-hand side (RHS):
RHS = 1 - 2cos^2(x)
Recall that 1 - cos^2(x) = sin^2(x), which means that 2cos^2(x) = 2(1 - sin^2(x)).
Substituting this into the RHS:
RHS = 1 - 2cos^2(x)
= 1 - 2(1 - sin^2(x))
= 1 - 2 + 2sin^2(x)
= -1 + 2sin^2(x)
Now, we need to show that the LHS is equal to the RHS. To do this, we will show that LHS - RHS = 0.
LHS - RHS = 1 - (-1 + 2sin^2(x))
= 1 + 1 - 2sin^2(x)
= 2 - 2sin^2(x)
We can rewrite 2 - 2sin^2(x) as 2(1 - sin^2(x)), which simplifies to 2cos^2(x).
Therefore, LHS - RHS simplifies to 2cos^2(x).
Since we know that LHS = 1 and LHS - RHS = 2cos^2(x), we can conclude that 1 = 2cos^2(x).
Hence, the given trigonometric identity (tan x + cot(-x))/(tan x - cot(-x)) = 1 - 2cos^2(x) has been proven.