Question

1)(sin2^альфа)/(1-sin^2альфа)*ctg^2альфа= 2)(sin альфа-cos альфа)^2+(cos альфа+sin альфа)^2=2

Answer 1

$\sf\displaystyle \frac{\sin^2\alpha}{1-\sin^2\alpha}\cdot ctg^2\alpha=\frac{\sin^2\alpha}{\cos^2\alpha}\cdot ctg^2\alpha=tg^2\alpha\cdot ctg^2\alpha=1$

$\tt\displaystyle (\sin\alpha-\cos\alpha)^2+(\cos\alpha+\sin\alpha)^2=2\\ \sin^2\alpha+\cos^2\alpha-2\cos\alpha\sin\alpha+\cos^2\alpha+\sin^2\alpha+2\cos\alpha\sin\alpha=2\\2cos^2\alpha+2\sin^2\alpha=2\\2=2$

Answer 2

Let's break down each of the given expressions step by step:

1. $\frac{\sin^2\alpha}{1-\sin^2\alpha}\cdot\cot^2\alpha$

We can manipulate the expression using trigonometric identities:

First, we know that $\cot\alpha = \frac{1}{\tan\alpha}$, and $\tan\alpha = \frac{\sin\alpha}{\cos\alpha}$.

$\cot\alpha = \frac{1}{\frac{\sin\alpha}{\cos\alpha}} = \frac{\cos\alpha}{\sin\alpha}$

Now, let's rewrite the original expression:

$\frac{\sin^2\alpha}{1-\sin^2\alpha}\cdot\cot^2\alpha = \frac{\sin^2\alpha}{1-\sin^2\alpha}\cdot\left(\frac{\cos\alpha}{\sin\alpha}\right)^2 = \frac{\sin^2\alpha}{1-\sin^2\alpha}\cdot\frac{\cos^2\alpha}{\sin^2\alpha}$

Now, we know the Pythagorean identity $\sin^2\alpha + \cos^2\alpha = 1$, which implies $\cos^2\alpha = 1 - \sin^2\alpha$. We can use this to simplify further:

$\frac{\sin^2\alpha}{1-\sin^2\alpha}\cdot\frac{\cos^2\alpha}{\sin^2\alpha} = \frac{\sin^2\alpha}{1-\sin^2\alpha}\cdot\frac{1-\sin^2\alpha}{\sin^2\alpha} = \frac{\sin^2\alpha(1-\sin^2\alpha)}{(1-\sin^2\alpha)\sin^2\alpha}$

Now, the term $(1-\sin^2\alpha)$ cancels out:

$\frac{\sin^2\alpha(1-\sin^2\alpha)}{(1-\sin^2\alpha)\sin^2\alpha} = \frac{\sin^2\alpha}{\sin^2\alpha} = 1$

So, the first expression simplifies to $1$.

2. $(\sin\alpha-\cos\alpha)^2 + (\cos\alpha+\sin\alpha)^2$

Let's expand and simplify this expression:

$(\sin\alpha-\cos\alpha)^2 = \sin^2\alpha - 2\sin\alpha\cos\alpha + \cos^2\alpha$

$(\cos\alpha+\sin\alpha)^2 = \cos^2\alpha + 2\sin\alpha\cos\alpha + \sin^2\alpha$