Does anyone know how to verify (1-tan^2x)/(1-cot^2x) = (1-sec^2x)?

L.H.S.
= (1 - tan²x) / (1 - cot²x)
= (1 - tan²x) / [1 - (1/tan²x)]
= (1 - tan²x) / [(tan² - 1)/tan²x]
= (1 - tan²x) / [tan²x/(tan² - 1)]
= -(tan²x - 1) / [tan²x/(tan² - 1)]
= -tan²x
= -(sec²x - 1)
= 1 - sec²x
= R.H.S.

Since L.H.S. = R.H.S.
(1 - tan²x) / (1 - cot²x) = 1 - sec²x

i) Multiply both numerator and denominator by tan²x;
so left side = tan²x(1 - tan²x)/(tan²x - 1) = -tan²x(1 - tan²x)/(1 - tan²x) = -tan²x

ii) By identity sec²x - tan²x = 1
So, -tan²x = 1 - sec²x

Thus (1 - tan²x)/(1 - cot²x) = -tan²x = 1 - seec²x = Right side [Proved]

NOTE: This is valid only when denominator is not equal to zero.
Or, domain of the given one is "x not equal to (nπ ± π/4), where n is an integer"

= [1 - tan²(x)] / [1 - cot²(x)] → you know that: tan(x) = sin(x)/cos(x)

= [1 - {sin(x)/cos(x)}²] / [1 - cot²(x)] → you know that: cot(x) = cos(x)/sin(x)

= [1 - {sin(x)/cos(x)}²] / [1 - {cos(x)/sin(x)}²]

= [{cos²(x)/cos²(x)} - {sin²(x)/cos²(x)}] / [{sin²(x)/sin²(x)} - {cos²(x)/sin²(x)}]

= [{cos²(x) - sin²(x)}/cos²(x)] / [{sin²(x) - cos²(x)}/sin²(x)]

= [{cos²(x) - sin²(x)}/cos²(x)] * [sin²(x)/{sin²(x) - cos²(x)}]

= [sin²(x).{cos²(x) - sin²(x)}] / [cos²(x).{sin²(x) - cos²(x)}] → recall: sin²(x) = 1 - cos²(x)

= [{1 - cos²(x)}.{cos²(x) - sin²(x)}] / [cos²(x).{sin²(x) - cos²(x)}] → recall: sin²(x) = 1 - cos²(x)

= [{1 - cos²(x)}.{cos²(x) - 1 + cos²(x)}] / [cos²(x).{1 - cos²(x) - cos²(x)}]

= [{1 - cos²(x)}.{2.cos²(x) - 1}] / [cos²(x).{1 - 2.cos²(x)}]

= - [{1 - cos²(x)}.{2.cos²(x) - 1}] / [cos²(x).{2.cos²(x) - 1}]

= - [1 - cos²(x)] / cos²(x)

= [cos²(x) - 1] / cos²(x)

= [cos²(x)/cos²(x)] - [1/cos²(x)]

= 1 - [1/cos²(x)]

= 1 - [1/cos(x)]² → recall: 1/cos(x) = sec(x)

= 1 - sec²(x)

Hint cot Θ = 1 / tan Θ