Prove the following triginmetric identity (√1-cosx/√1+cosx)= cosecx - cotx

Answer 1

Begin by multiplying both numerator and denominator by √(1 - cos(x)) such that

(1) √(1 - cos(x))/√(1 + cos(x)) = [√(1 - cos(x))/√(1 + cos(x))] * [√(1 - cos(x))/√(1 - cos(x))]

(2) √(1 - cos(x))/√(1 + cos(x)) = (1 - cos(x))/√(1 - cos²(x))

(3) √(1 - cos(x))/√(1 + cos(x)) = (1 - cos(x))/√(sin²(x))

(4) √(1 - cos(x))/√(1 + cos(x)) = (1 - cos(x))/sin(x)

(5) √(1 - cos(x))/√(1 + cos(x)) = 1/sin(x) - cos(x)/sin(x)

(6) √(1 - cos(x))/√(1 + cos(x)) = csc(x) - cot(x)

In step 2 -> 3 we use the identity 1 - cos²(x) = sin²(x) (Derivable from the fact that sin²(x) + cos²(x) = 1).

In step 3 -> 4 we simply change √(sin²(x)) to sin(x).

In step 4 -> 5 we distribute the 1/sin(x) to each term in the parentheses.

In step 5 -> 6 we use two more trig identities to simplify and yield the final result.