Question

Ctg(b-a) (tga-tgb) +tgatgb+cos2y​

Answer 1

Ответ:

$\cos 2y-1$

Объяснение:

$\cot(b-a)(\tan a-\tan b)+\tan a*\tan b+\cos 2y=\\=\frac{\cos(b-a)}{\sin(b-a)}(\frac{\sin a}{\cos a}-\frac{\sin b}{\cos b})+\frac{\sin a\sin b}{\cos a\cos b}+\cos 2y=\\=\frac{\cos a\cos b+ \sin a\sin b}{\sin b \cos a-\cos b\sin a}\frac{\sin a\cos b-\cos a\sin b}{\cos a\cos b}+\frac{\sin a\sin b}{\cos a\cos b}+\cos 2y=\\=-\frac{\cos a\cos b+ \sin a\sin b}{\cos a\cos b}+\frac{\sin a\sin b}{\cos a\cos b}+\cos 2y=\\=-\frac{\cos a\cos b}{\cos a\cos b}+\cos 2y=\cos 2y-1$

Answer 2

The expression you provided appears to be a combination of trigonometric functions. Let's break it down step by step:

1. Ctg(b-a): This represents the cotangent of the quantity (b - a). Cotangent is the reciprocal of the tangent function and can be calculated as 1/tan(b - a).

2. (tga - tgb): This is the difference between two tangent functions, tga and tgb.

3. tgatgb: This appears to be the product of the tangent of 'a' and the tangent of 'b'.

4. cos2y: This is the cosine of 2y.

To simplify the expression further, you can apply trigonometric identities and combine like terms. For example, you can rewrite cos(2y) in terms of other trigonometric functions using the double-angle identity for cosine:

cos(2y) = 2cos^2(y) - 1

So, the expression can be written as:

(1/tan(b - a)) * (tga - tgb) + (tga * tgb) + 2cos^2(y) - 1

You can further simplify or manipulate the expression based on specific values or requirements.