proof that tanx + secx -1/tanx - secx+1 = 1+sinx/cosx
The way you typed it, it is false.
It needs brackets, but the number of variations to
place those brackets is too great for me to start guessing
what you meant.
e.g. did you mean :
tanx + (secx -1)/(tanx - secx)+1 = (1+sinx)/cosx
or
(tanx + secx -1)/(tanx - secx+1) = 1+sinx/cosx
or
....
To prove the given trigonometric identity:
tan(x) + sec(x) - 1 / tan(x) - sec(x) + 1 = 1 + sin(x) / cos(x)
We first need to simplify both sides of the equation.
Let's simplify the left side of the equation:
tan(x) + sec(x) - 1 / tan(x) - sec(x) + 1
To simplify this, we can combine the fractions by finding a common denominator.
The common denominator is (tan(x) - sec(x) + 1). Multiply each fraction by this common denominator:
(tan(x) + sec(x) - 1) * (tan(x) - sec(x) + 1) / (tan(x) - sec(x) + 1) + 1 * (tan(x) - sec(x) + 1) / (tan(x) - sec(x) + 1)
Expanding the fractions, we get:
[tan(x)(tan(x) - sec(x) + 1) + sec(x)(tan(x) - sec(x) + 1) - (tan(x) - sec(x) + 1)] / (tan(x) - sec(x) + 1)
Now, simplify the numerator:
[tan^2(x) - tan(x)sec(x) + tan(x) + sec(x)tan(x) - sec^2(x) + sec(x) - tan(x) + sec(x) - 1] / (tan(x) - sec(x) + 1)
Combining like terms, we get:
[tan^2(x) - sec^2(x) + 2tan(x) + 2sec(x) - 1] / (tan(x) - sec(x) + 1)
Now, let's focus on the right side of the equation:
1 + sin(x) / cos(x)
We can simplify the right side by multiplying both the numerator and denominator by cos(x):
[cos(x) + sin(x)] / cos(x)
Now, let's simplify the left side further:
[tan^2(x) - sec^2(x) + 2tan(x) + 2sec(x) - 1] / (tan(x) - sec(x) + 1) = [cos(x) + sin(x)] / cos(x)
To prove the given trigonometric identity, we need to show that the left side is equal to the right side.
[tan^2(x) - sec^2(x) + 2tan(x) + 2sec(x) - 1] / (tan(x) - sec(x) + 1) = [cos(x) + sin(x)] / cos(x)
To continue with the proof, we can simplify the left side of the equation. However, it appears that there may be an error in the given equation, as the two sides do not seem to be equal. Please double-check the expression you provided, and let me know if there is a correction to be made.