Question

([email protected][email protected])^2-(ctg-tg)^2

Answer 1

Ответ:

[email protected]^[email protected]^[email protected]@-([email protected]^[email protected]^[email protected]@= [email protected]^[email protected]^[email protected]@[email protected]^2 [email protected]^[email protected]@= [email protected]@= 4 ([email protected]×[email protected]=1 по формуле)

Answer 2

It seems like you've provided a mathematical expression: $(\csc \theta - \sec \theta)^2 - (\cot \theta - \tan \theta)^2$. Let's simplify this expression step by step.

Recall the trigonometric identities:

1. $\csc \theta = \frac{1}{\sin \theta}$
2. $\sec \theta = \frac{1}{\cos \theta}$
3. $\cot \theta = \frac{1}{\tan \theta}$
4. $\tan \theta = \frac{\sin \theta}{\cos \theta}$

Using these identities, we can rewrite the expression:

$$\begin{align*} & (\csc \theta - \sec \theta)^2 - (\cot \theta - \tan \theta)^2 \\ &= \left(\frac{1}{\sin \theta} - \frac{1}{\cos \theta}\right)^2 - \left(\frac{1}{\tan \theta} - \frac{\sin \theta}{\cos \theta}\right)^2 \\ &= \left(\frac{\cos \theta - \sin \theta}{\sin \theta \cos \theta}\right)^2 - \left(\frac{\cos \theta - \sin^2 \theta}{\sin \theta \cos \theta}\right)^2 \\ &= \frac{(\cos \theta - \sin \theta)^2 - (\cos \theta - \sin^2 \theta)^2}{\sin^2 \theta \cos^2 \theta} \\ &= \frac{\cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta - \cos^2 \theta + 2 \cos \theta \sin^2 \theta - \sin^4 \theta}{\sin^2 \theta \cos^2 \theta} \\ &= \frac{2 \cos \theta \sin^2 \theta - \sin^4 \theta}{\sin^2 \theta \cos^2 \theta} \\ &= \frac{\sin^2 \theta (2 \cos \theta - \sin^2 \theta)}{\sin^2 \theta \cos^2 \theta} \\ &= \frac{2 \cos \theta - \sin^2 \theta}{\cos^2 \theta}. \end{align*}$$

So, the simplified expression is $\frac{2 \cos \theta - \sin^2 \theta}{\cos^2 \theta}$.