verify that the equation is an identity:
(1-cosX)/(1+cosX) = (cscX-cotX)^2
LS = (1- cosx)/(1 + cosx) [(1-cox)/(1-cosx)]
= (1-cosx)^2/(1-cos^2x)
RS = (1/sinx - cosx/sinx)^2
= ((1-cosx)/sinx)^2
= (1-cosx)^2 /(sin^2x)
= (1-cosx)^2/(1-cos^2)
= LS
thanks. How would you know how to do similar problems? Where should i start when verifying trig equations?
Usually start with the complicated looking side and try to reduce to the simpler side.
In this case I worked on both sides.
Also I often change everything to sines and cosines, unless I recognize a popular identity.
How did I know where to start??
Let's say doing this for 35 years and doing thousands of these could be the reason, lol
To verify whether the equation is an identity, we need to simplify both sides of the equation and check if they are equal for all values of X.
Starting with the left side of the equation:
(1 - cosX) / (1 + cosX)
To simplify this expression, we will multiply the numerator and denominator by the conjugate of the denominator, which is (1 - cosX):
[(1 - cosX)(1 - cosX)] / [(1 + cosX)(1 - cosX)]
(1 - 2cosX + cos^2X) / (1 - cos^2X)
(1 - 2cosX + cos^2X) / sin^2X
Now, let's simplify the right side of the equation:
(cscX - cotX)^2
Expanding the square:
(cscX - cotX)(cscX - cotX)
(csc^2X - 2cscXcotX + cot^2X)
(1/sin^2X - 2cosX/sin^2X + cos^2X/sin^2X)
To determine if these two expressions are equal, we need to manipulate them further.
First, let's write the trigonometric identities for csc and cot in terms of sin and cos:
cscX = 1/sinX
cotX = cosX/sinX
Now we can substitute these identities into the right side of the equation:
(1/sin^2X - 2cosX/sin^2X + cos^2X/sin^2X)
(1/sin^2X - 2cosX/sin^2X + (1 - sin^2X)/sin^2X)
(1 - 2cosX + 1 - sin^2X) / sin^2X
(2 - 2cosX - sin^2X) / sin^2X
(2 - (2cosX + sin^2X)) / sin^2X
Now, let's go back to the left side of the equation and simplify it further:
(1 - 2cosX + cos^2X) / sin^2X
At this point, we can see that the numerator on both sides of the equation is the same:
(1 - 2cosX + cos^2X)
So, we can rewrite the equation:
(2 - (2cosX + sin^2X)) / sin^2X = (2 - (2cosX + sin^2X)) / sin^2X
Since the numerators are the same and the denominators are the same, we have shown that the equation is an identity.
Therefore, the equation (1 - cosX) / (1 + cosX) = (cscX - cotX)^2 is indeed an identity.