Question

Ctg2a-cos2a/tg2a-sin2a

Answer 1

$\displaystyle \frac{ctg2a-cos2a}{tg2a-sin2a}= \frac{ \frac{cos2a}{sin2a}- \frac{cos2a*sin2a}{sin2a}}{ \frac{sin2a}{cos2a}- \frac{sin2a*cos2a}{cos2a}}=\\\\= \frac{cos2a(1-sin2a)}{sin2a}: \frac{sin2a(1-cos2a)}{cos2a}=\\\\= \frac{cos^22a(1-sin2a)}{sin^22a(1-cos2a)}=ctg^22a* \frac{cos^2a+sin^2a-2sina*cosa}{cos^2+sin^2a-(cos^2a-sin^2a)}=\\\\=ctg^22a \frac{(cosa-sina)^2}{2sin^2a}$

более эту дробь не упростить

но в аналогичных заданиях требовалось доказать что это выражение будет больше 0
Проверим это: 
т.к. выражение представляем собой произведение квадратов чисел то его общее значение действительно всегда будет больше 0

Answer 2

The expression you provided is:

(ctg^2a - cos^2a) / (tg^2a - sin^2a)

Let's simplify it step by step:

First, we can rewrite ctg^2a as (1/tan^2a), cos^2a as (1 - sin^2a), and tg^2a as (sin^2a / cos^2a):

[(1/tan^2a) - (1 - sin^2a)] / [(sin^2a / cos^2a) - sin^2a]

Now, let's simplify further:

[(1/tan^2a) - 1 + sin^2a] / [(sin^2a / cos^2a) - sin^2a]

Next, we'll find a common denominator for both fractions:

[(1 - tan^2a + tan^2a * sin^2a) / tan^2a] / [(sin^2a - sin^4a) / cos^2a]

Simplifying the numerator:

[1 / tan^2a] / [(sin^2a - sin^4a) / cos^2a]

Inverting the denominator by multiplying it with (cos^2a / 1):

[1 / tan^2a] * [cos^2a / (sin^2a - sin^4a)]

Now, we can simplify further by canceling out common factors:

cos^2a / (tan^2a * (sin^2a - sin^4a))

Finally, using the identity tan^2a = sin^2a / cos^2a:

cos^2a / ((sin^2a / cos^2a) * (sin^2a - sin^4a))

cos^2a / (sin^2a - sin^4a)

That's the simplified form of the expression ctg^2a - cos^2a / tg^2a - sin^2a.