trig help please?

Question

Prove the identity.

(1 / (secΘ - 1)) - (1 / (secΘ + 1)) = 2 cot²Θ

can anyone please help me prove this identity?

thank you

Answer 1

1/(secΘ - 1) - 1/(secΘ + 1)

find the LCD
(secΘ+1) / [(secΘ-1)(secΘ+1)] - (secΘ-1)/[(secΘ-1)(secΘ+1)]

simplify
(secΘ + 1) / (sec^2Θ - 1) - (secΘ - 1) / (sec^2Θ - 1)

simplify
(secΘ + 1 - secΘ + 1) / (sec^2Θ - 1)

simplify
2 / (sec^2Θ - 1)

trig identity
sec^2Θ = 1 + tan^2Θ

substitute
2 / (1 + tan^2Θ - 1)

simplify
2 / (tan^2Θ)

2 * 1/tan^2Θ

2 cot^2Θ

Answer 2

Get a common denominator on the left and combine the two fractions. the numerator will equal 2. the denominator will be of the form (a-b) (a+b)= a^2 - b^2, which in this case is

[sec theta]^2-1 = [tan theta]^2

and of course tan in the denominator is the same as cot in the numerator.

Answer 3

start with left hand side

=2/(sec^2-1)
but denominator equals tan^2
=2/tan^2
=2cot^2

Answer 4

doing the left side
1/(sec x -1) - 1/(sec x + 1)

((sec x +1)- (sec x-1))/(sec^2 x-1)

(2/sec^2 x -1)
(2/tan^2 x ) = 2 * (1/tan^2 x) = 2cot^2 x