cosx/1-tanx + sinx/1-cotx = sinx =cosx
Please check if there should be two equal signs in the equation, and also if the 1-tanx and 1-cotx should have been in parentheses.
Never mind!
See next post:
http://www.jiskha.com/display.cgi?id=1243981839
To solve the equation cos(x)/(1 - tan(x)) + sin(x)/(1 - cot(x)) = sin(x) + cos(x), we can simplify each term step by step.
Let's start with the left side of the equation:
cos(x)/(1 - tan(x)) + sin(x)/(1 - cot(x))
We know that tan(x) is equal to sin(x)/cos(x) and cot(x) is equal to cos(x)/sin(x). Substituting these values, we have:
cos(x)/(1 - sin(x)/cos(x)) + sin(x)/(1 - cos(x)/sin(x))
Next, we need to simplify the fractions:
cos(x)/((cos(x) - sin(x))/cos(x)) + sin(x)/((sin(x) - cos(x))/sin(x))
Next, we can simplify by multiplying each fraction by the reciprocal of the denominator:
cos(x) * (cos(x)/ (cos(x) - sin(x))) + sin(x) * (sin(x)/ (sin(x) - cos(x)))
Simplifying further, we get:
cos^2(x)/ (cos(x) - sin(x)) + sin^2(x)/ (sin(x) - cos(x))
Since cos^2(x) = 1 - sin^2(x), we can substitute this value:
(1 - sin^2(x))/ (cos(x) - sin(x)) + sin^2(x)/ (sin(x) - cos(x))
Now we can combine the two fractions by finding the common denominator:
((1 - sin^2(x))(sin(x) - cos(x))) + (sin^2(x))(cos(x) - sin(x))) / ((cos(x) - sin(x))
((sin(x) - cos(x) - sin^3(x) + sin^2(x)*cos(x)) + (cos(x) - sin(x))*sin^2(x)) / (cos(x) - sin(x))
Now we can see that many terms cancel out:
(sin(x) - cos(x) - sin^3(x) + sin^2(x)*cos(x) + cos(x)*sin^2(x) - sin(x)*sin^2(x)) / (cos(x) - sin(x))
(- cos(x) - sin^3(x) + sin^2(x)*cos(x) + sin(x)*cos(x) - sin^3(x)) / (cos(x) - sin(x))
Rearranging the terms, we have:
(- cos(x) - 2sin^3(x) + 2sin^2(x)*cos(x)) / (cos(x) - sin(x))
So the simplified expression on the left side of the equation is:
(- cos(x) - 2sin^3(x) + 2sin^2(x)*cos(x)) / (cos(x) - sin(x))
Since the simplified expression on the left side is not equal to sin(x) + cos(x), we can conclude that the original equation cos(x)/(1 - tan(x)) + sin(x)/(1 - cot(x)) = sin(x) + cos(x) is not true for all values of x.