Question

1)tg^2(270°+a)2)cos^2(270°-a)3)sin^2(180°-x)+sin^2(270°-x)4)cos^2(180°+x)+cos^2(90°+x)

Answer 1

1) tg²(270°+a) = ctg² a
2) cos²(270°-a) = sin² a
3) sin²(180°-x) +sin²(270°-x) = sin²x+cos²x=1
4) cos²(180°+x) +cos²(90°+x) = cos²x+sin²x=1

Answer 2

1. We know that the tangent function has a period of 180 degrees, so tg(270° + a) is equivalent to tg(90° + a), which has the same value as -cot(a). Therefore, tg^2(270° + a) = (-cot(a))^2 = cot^2(a).

2. We know that cos(270° - a) is equivalent to sin(a), so cos^2(270° - a) = sin^2(a).

3. Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can rewrite sin^2(180° - x) + sin^2(270° - x) as (cos^2(x) + sin^2(x)) + (sin^2(x) + cos^2(x)) - sin^2(x) = 2cos^2(x).

4. Similarly, using the identity cos^2(x) + sin^2(x) = 1, we can rewrite cos^2(180° + x) + cos^2(90° + x) as (sin^2(x) + cos^2(x)) + (cos^2(x) + sin^2(x)) - cos^2(x) = 2sin^2(x).

Therefore:

1. tg^2(270°+a) = cot^2(a)
2. cos^2(270°-a) = sin^2(a)
3. sin^2(180°-x) + sin^2(270°-x) = 2cos^2(x)
4. cos^2(180°+x) + cos^2(90°+x) = 2sin^2(x)