Trigonometry Identity II

Question

Please fully complete this identity, step-by-step with as much detail as necessary!

(Note: 'θ' (theta) just means angle. It can be replaced with 'x'.)

(1 ÷ cos θ + 1) - (1 ÷ cos θ - 1) = (2 ÷ sin² θ)

Answer 1

Please use parenthesis, so that we can tell what expressions are in the denominator, and which are not.

Also, after awhile, I end up eliminating the argument of the trig function altogether.

Forthwith, it is sin (x) = sin x = sin and

cos (x) = cos x = cos

1 / (cos x +1) - 1/(cos x - 1) = <--- the common denominator is (cos x + 1)(cos x - 1) =

(cos x)^2 - 1

(cos x - 1)/ [(cos x)^2 - 1 ] - (cos x + 1)/ [(cos x)^2 - 1 ]

(cos x - 1 - (cos x + 1) / [(cos x)^2 - 1 ]

(cos x - 1 - cos x - 1) / [(cos x)^2 - 1 ]

-2/[(cos x)^2 - 1 ]

BUT.... sin^2 + cos^2 = 1

sin^2 = 1 - cos^2

-sin^2 = cos^2 - 1 <--- this is the denominator

So

-2/ [(cos x)^2 - 1 ] =

-2/ (-sin^2) =

2/ sin^2

which completes the proof

got it?