1/tanx-secx+ 1/tanx+secx=-2tanx
so this is what I did:
=tanx+secx+tanx-secx
=(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx)
=sinx/cosx+ sinx /cosx= -2tanxI
but I know this can't be correct because what I did doesn't end as a negatvie:( please help
To solve the equation 1/tan(x) - sec(x) + 1/tan(x) + sec(x) = -2tan(x), we can simplify the expression first.
1/tan(x) is the same as cos(x)/sin(x) and sec(x) is 1/cos(x), so we can rewrite the equation as:
cos(x)/sin(x) - 1/cos(x) + cos(x)/sin(x) + 1/cos(x) = -2tan(x).
Now, let's find a common denominator to combine the fractions:
(sin(x)*cos(x) - 1 + cos(x)*sin(x) + 1) / (sin(x)*cos(x)) = -2tan(x).
Simplifying further:
(2*sin(x)*cos(x)) / (sin(x)*cos(x)) = -2tan(x).
The sin(x)*cos(x) terms cancel out leaving:
2 = -2tan(x).
Divide both sides of the equation by -2:
2 / -2 = tan(x).
Simplifying gives:
-1 = tan(x).
Therefore, the value of x that satisfies the equation is any angle whose tangent is -1. This includes angles such as -45 degrees or 225 degrees, among others.