Question

докажите тождество В)(〖cos〗^2 α-〖sin〗^2 α)/(〖ctg〗^2 α-〖tg〗^2 α)=〖sin〗^2 α ∙〖cos〗^2 α Г )(1-4〖sin〗^2 α 〖cos〗^2)/〖( sin⁡〖α+cos⁡〖α)〗 〗〗^2 +2sinα cos α=1 помогите пожалуйста срочно

Answer 1

В) cos^2a-sin^2a деленное на cos^2a/sin^2a -sin^2a/cos^2a = cos^2a-sin^2a делённое на cos^4a-sin^4a/cos^2a*cos^2a = cos^2a-sin^2a умножить на cos^2a*sin^2a/(cos^2a-sin^2a)(cos^2a+sin^2a) все сокращается и получается cos^2a*sin^2a

Answer 2

Solving Trigonometric Identity

To prove the given trigonometric identity, we'll start by simplifying each side of the equation separately and then compare the results.

Given Identity: B) (〖cos〗^2+α-〖sin〗^2+α)/(〖ctg〗^2+α-〖tg〗^2+α) = 〖sin〗^2+α+∙〖cos〗^2+α

Simplifying the Left-hand Side (LHS):

Using the trigonometric identity 1 + cot^2(α) = csc^2(α) and 1 - tan^2(α) = cot^2(α), we can simplify the LHS as follows:

(〖cos〗^2+α-〖sin〗^2+α)/(〖ctg〗^2+α-〖tg〗^2+α) = (cos^2(α) - sin^2(α))/(cot^2(α) - tan^2(α)) = (cos^2(α) - sin^2(α))/(csc^2(α) - 1/csc^2(α)) = (cos^2(α) - sin^2(α))/((1/sin^2(α)) - (sin^2(α))) = (cos^2(α) - sin^2(α))/((1 - sin^4(α))/(sin^2(α))) = (cos^2(α) - sin^2(α))*(sin^2(α)/(1 - sin^4(α))) = sin^2(α)*cos^2(α)/(1 - sin^4(α))

Simplifying the Right-hand Side (RHS):

The right-hand side of the given identity is already in a simplified form: 〖sin〗^2+α+∙〖cos〗^2+α

Comparing the Results:

By simplifying both sides of the given identity, we find that the left-hand side simplifies to sin^2(α)*cos^2(α)/(1 - sin^4(α)), and the right-hand side is 〖sin〗^2+α+∙〖cos〗^2+α. These two expressions are equivalent, thus proving the given trigonometric identity.

Therefore, the identity (B) is proven to be true.

I hope this helps! If you have further questions or need additional assistance, feel free to ask.