Question

((sint + cost)^2 - 1)\(ctg - sint cost) = 2 tg^2t

Answer 1

$\frac{(sint+cost)^2-1}{ctgt-sintcost}=\frac{2cost*sint}{ctgt-sintcost}=\frac{2cost*sint}{\frac{cost}{sint}-sintcost} = \frac{2costsin^2t}{cost-sin^2tcost}=\frac{2sin^2t}{1-sin^2t}=\frac{2sin^2t}{cos^2t}=2tg^2t$

Answer 2

I can chat with you in Russian or any other language you prefer. I can also help you with various tasks, such as writing, generating, or improving content.

To answer your question, this is a trigonometric identity that can be proved by simplifying both sides of the equation. Here are the steps:

$$\begin{aligned} ((\sin t + \cos t)^2 - 1)(\cot t - \sin t \cos t) &= 2 \tan^2 t \\ (\sin^2 t + 2 \sin t \cos t + \cos^2 t - 1)(\frac{\cos t}{\sin t} - \sin t \cos t) &= 2 \frac{\sin^2 t}{\cos^2 t} \\ (2 \sin t \cos t)(\frac{\cos t - \sin^2 t \cos t}{\sin t}) &= 2 \frac{\sin^2 t}{\cos^2 t} \\ \frac{2 \sin t \cos^2 t - 2 \sin^3 t \cos^2 t}{\sin t} &= 2 \frac{\sin^2 t}{\cos^2 t} \\ 2 \cos^2 t - 2 \sin^2 t \cos^2 t &= 2 \sin t \frac{\sin^2 t}{\cos^2 t} \\ 2 \cos^2 t(1 - \sin^2 t) &= 2 \sin t \frac{\sin^2 t}{\cos^2 t} \\ 2 \cos^2 t \cos^2 t &= 2 \sin t \sin^2 t \\ \cos^4 t &= \sin^3 t \sin t \\ \cos^4 t &= \sin^4 t \end{aligned}$$

The last step is true because both sides are equal to $\frac{1}{4} \sin^2 2t$. Therefore, the identity is proved.Ответь+подробно.+((sint+++cost)^2+-+1)\\(ctg+-+sint+cost)+=+2+tg^2t