square root of (1-cosx/1+cosx) = cscx - cotx
prove the identity
To prove the identity √(1-cosx/1+cosx) = cscx - cotx, we need to simplify both sides of the equation and show that they are equal.
First, let's simplify the left-hand side (LHS):
√(1-cosx/1+cosx)
We can simplify the fraction by multiplying both the numerator and the denominator by (1-cosx) to get rid of the division:
√((1-cosx)(1-cosx)/(1+cosx)(1-cosx))
Expanding the numerator and the denominator:
√(1-2cosx+cos^2x)/(1-cos^2x)
Since cos^2x + sin^2x = 1 (a fundamental trigonometric identity), we can rewrite the denominator as 1 - cos^2x:
√(1-2cosx+cos^2x)/(sin^2x)
Taking the square root of the numerator and simplifying:
√((1-cosx)^2)/(sin^2x)
Simplifying the square root of the numerator:
(1-cosx)/(sinx)
Now let's simplify the right-hand side (RHS):
cscx - cotx
Using the definitions of cscx and cotx:
1/sinx - cosx/sinx
Combining the fractions:
(1 - cosx)/sinx
Now we can see that the simplified LHS and RHS are the same:
(1-cosx)/(sinx) = (1-cosx)/(sinx)
Therefore, we have successfully proven the identity √(1-cosx/1+cosx) = cscx - cotx.