verify the identity
(1-cosx)/(sinx)+(sinx)/(1-cosx)=2cscx
Thank you!
LS = (1-cosx)/(sinx)+(sinx)/(1-cosx)
= ( (1-cosx)(1-cosx) + sinx(sinx) )/(sinx(1-cosx))
= (1 - 2cosx + cos^2 x + sin^2 x)/(sinx(1-cosx))
= (1 - 2cosx + 1)/(sinx(1-cosx)
= 2(1-cosx)/(sinx(1-cosx))
= 2/sinx
= 2 cscx
= RS
To verify the given identity, we need to simplify the left-hand side (LHS) of the equation and show that it is equal to the right-hand side (RHS).
Let's start by simplifying the LHS of the equation:
(1 - cos(x))/(sin(x)) + (sin(x))/(1 - cos(x))
To combine the two fractions with the common denominator, we multiply the first fraction by (1 - cos(x))/(1 - cos(x)) and the second fraction by (sin(x))/(sin(x)):
[(1 - cos(x))(1 - cos(x))] / [(sin(x))(1 - cos(x))] + [(sin(x))(sin(x))] / [(1 - cos(x))(sin(x))]
Expanding and simplifying each term:
[(1 - 2cos(x) + cos^2(x)) + sin^2(x)] / [(sin(x) - sin(x)cos(x))]
Combining like terms:
[1 - 2cos(x) + cos^2(x) + sin^2(x)] / [sin(x) - sin(x)cos(x)]
Using the Pythagorean identity (sin^2(x) + cos^2(x) = 1), we can simplify further:
[1 - 2cos(x) + 1] / [sin(x) - sin(x)cos(x)]
Simplifying the numerator:
[2 - 2cos(x)] / [sin(x) - sin(x)cos(x)]
Factoring out a 2 from the numerator:
2(1 - cos(x)) / [sin(x) - sin(x)cos(x)]
Using the identity sin(x) = 1/csc(x) and cos(x) = 1/sec(x), we can rewrite the expression:
2(1 - 1/sec(x)) / [1/csc(x) - 1/csc(x) * 1/sec(x)]
Simplifying further:
2(1 - 1/sec(x)) / [1/csc(x) - 1/sec(x) * csc(x)]
Applying the reciprocal identities (sec(x) = 1/cos(x) and csc(x) = 1/sin(x)):
2(1 - cos(x)/1) / [1/sin(x) - 1/cos(x) * 1/sin(x)]
Simplifying:
2(1 - cos(x)) / [1 - 1/cos(x)]
Multiplying the denominator by cos(x):
2(1 - cos(x)) / [cos(x) - 1]
Finally, simplifying the expression further:
2(-cos(x)) / (cos(x) - 1)
Now, we can see that the simplified expression matches the RHS of the equation, which is 2csc(x). Therefore, the given identity:
(1 - cos(x))/(sin(x)) + (sin(x))/(1 - cos(x)) = 2csc(x)
has been verified.