Question

1)sin^2α+1/cos^2α-1+tg*ctg 2)sin(π/2+α)tg(π-α)/cos(3/2π-α)ctg(π+α)

Answer 1

1)(sin²α+1)/(cos²α-1)+tgα*ctgα=(sin²α+1)/(-sin²α)+1=-1-(1/(sin²α))+1=-(1/(sin²α))

2)sin(π/2+α)tg(π-α)/cos(3/2π-α)ctg(π+α)=cosα*(-tgα) /(-sinα)ctgα=

sinα/cosα=tgα

Answer 2

Let's simplify the given expressions:

1. $\sin^2(\alpha) + \frac{1}{{\cos^2(\alpha) - 1}} + \tan(\alpha) \cot(\alpha)$

Starting with the second term, we can rewrite it as:

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

Using the identity $\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta)$, we get:

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

Now, let's simplify the third term:

$\tan(\alpha) \cot(\alpha)$

Using the identity $\cot(\theta) = \frac{1}{{\tan(\theta)}}$, we can rewrite it as:

$\tan(\alpha) \cot(\alpha) = \tan(\alpha) \cdot \frac{1}{{\tan(\alpha)}} = 1$

Now, substituting these simplifications back into the original expression, we have:

$\sin^2(\alpha) + \frac{1}{{\cos(2\alpha)}} + 1$

2. $\sin\left(\frac{\pi}{2} + \alpha\right) \cdot \tan\left(\pi - \alpha\right) \div \cos\left(\frac{3}{2}\pi - \alpha\right) \cdot \cot\left(\pi + \alpha\right)$

Let's simplify each term individually:

$\sin\left(\frac{\pi}{2} + \alpha\right) = \cos(\alpha)$

$\tan\left(\pi - \alpha\right) = -\tan(\alpha)$

$\cos\left(\frac{3}{2}\pi - \alpha\right) = -\sin(\alpha)$

$\cot\left(\pi + \alpha\right) = -\tan(\alpha)$

Now, substituting these simplifications back into the original expression, we have:

$\cos(\alpha) \cdot (-\tan(\alpha)) \div (-\sin(\alpha)) \cdot (-\tan(\alpha))$

Simplifying further:

$\cos(\alpha) \cdot \tan(\alpha) \div \sin(\alpha) \cdot \tan(\alpha)$

Now, we can cancel out the common factors of $\tan(\alpha)$:

$\frac{{\cos(\alpha)}}{{\sin(\alpha)}}$

This is equivalent to the cotangent of $\alpha$, so the simplified expression is:

$\cot(\alpha)$