Prove 2cotx 2cscxSolutionThere is some mistake So I am con

Prove 2cotx = 2cscx

Solution

There is some mistake . So I am considering it as

[(1+cosx)/sinx] + [sinx/(1-cosx)] = 2cscx + 2cotx

I\'ll start with the right side of the equation:

2cscx + 2cotx = 2/sinx + 2/tanx = 2/sinx + 2/sinx/cosx =

2/sinx + 2cosx/sinx = (2 + 2cosx)/sinx = 2(1 +cosx)/sinx

Now, remember that answer and solve the left side

[(1 + cosx)/sinx] + [sinx/(1 - cosx)] multiply by common denominator

[(1 + cosx)(1 - cosx) + (sinx)(sinx)]/(sinx)(1 - cosx) =

[(1 - cosx^2) + (sinx^2)]/(sinx)(1 - cosx)

Now, we know (cosx^2 +sinx^2) = 1, so (1 - cosx^2) = sinx^2 and the reverse as well

[(sinx^2) + (sinx^2)]/(sinx)(1 - cosx) =

(2sinx^2)/(sinx)(1 - cosx) the sin\'s cancel =

(2sinx)/(1 - cosx)

Now, we multiply the fraction by 1 in the form of (1 + cosx)/(1 + cosx) to change the denominator

(2sinx)(1 + cosx)/(1 - cosx)(1 + cosx) use the rule two above =

(2sinx)(1 + cosx)/(1 - cosx^2) =

(2sinx)(1 + cosx)/(sinx^2) the sin\'s cancel =

2(1 + cosx)/(sinx)