Verify: -ln(1+cosθ)=ln(1-cosθ)-2ln|sinθ|?

Question

Also: 
1. Simplify (sec x - cos x)^2 
2. Simplify: (tan x + cot x)^2 - (tan x - cot x)^2

Answer 1

The equation is - ln(1 + cos θ) = ln(1 - cos θ) - 2ln|sin θ|.

Left hand side : - ln[1 + cosθ].

Apply power property of logarithm :  .

= ln[1 + cosθ] ^{- 1}

= ln{1/(1 + cosθ)}

Multiply numerator and denominator by (1 - cosθ).

= ln{(1 - cos θ)/[(1 + cosθ)(1 - cosθ)}

= ln[(1 - cos θ)/(1 - cos² θ)]

Pythagorean identity : sin² θ + cos² θ = 1.

= ln[(1 - cos θ)/sin² θ]

Apply quotient property logarithm: .

ln(1 - cos θ) - ln|sin² θ|

Apply power property of logarithm :  .

ln(1 - cos θ) - 2ln|sin θ|

= Right hand side.

• 1).

The expression is (sec x - cos x )² .

Square of the difference : (a - b)² = a² - 2ab + b² .

(sec x - cos x )²  = sec² x - 2sec x cos x + cos² x

Using reciprocal identity : sec x = 1/cos x.

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

= (1/cos² x) - 2 + cos² x

= (1 - 2cos² x + cos⁴ x)/cos² x

= (1 - cos² x)²/cos² x

Pythagorean identity : sin² x + cos² x = 1.

= (sin² x)²/cos² x

= sin⁴ x/cos² x

= (sin² x/cos² x) * sin² x

Trigonometric identity : tan x = sin x/cos x.

= tan² x sin² x.

Therefore, (sec x - cos x )² = tan² x sin² x.

• 2).

The expression is (tan x + cot x)² - (tan x - cot x)² .

(tan x + cot x)² - (tan x - cot x)²

Difference of squares : a² - b² = (a + b)(a - b).

= [(tan x + cot x) + (tan x - cot x)] * [(tan x + cot x) - (tan x - cot x)]

= [tan x + cot x + tan x - cot x] * [tan x + cot x - tan x + cot x]

= [tan x + tan x] * [cot x + cot x]

= [2tan x] * [2cot x]

= 4tan x cot x

Using reciprocal identity : cot x = 1/tan x.

= 4 tan x * (1/tan x)

= 4.

Therefore, (tan x + cot x)² - (tan x - cot x)² = 4.