81 номер ПОЖАЛУЙСТА ОЧЕНЬ СРОЧНО!!!!!!!!!!!!!!!!!!! ЗАРАНЕЕ СПАСИБО ОГРОМНОЕ

Question 1

81 номер
ПОЖАЛУЙСТА
ОЧЕНЬ СРОЧНО!!!!!!!!!!!!!!!!!!!
ЗАРАНЕЕ СПАСИБО ОГРОМНОЕ

Question 2

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

$\frac{sin^2 \beta - cos^2 \beta +1 }{sin^2 \beta } = \frac{sin^2 \beta + sin^2 \beta }{sin^2 \beta } = \frac{2sin^2 \beta }{sin^2 \beta } = 2$

$\frac{1}{1+tg^2 \beta } + \frac{1}{1+ctg^2 \beta }= \frac{1}{1+ \frac{sin^2 \beta }{cos^2 \beta } } + \frac{1}{1+ \frac{cos^2 \beta }{sin^2 \beta } }= \frac{1}{\frac{cos^ \beta +sin^2 \beta }{cos^2 \beta } } + \frac{1}{ \frac{sin^2 \beta +cos^2 \beta }{sin^2 \beta } }= \\ = \frac{1}{\frac{1 }{cos^2 \beta } } + \frac{1}{ \frac{1}{sin^2 \beta } }=cos^2 \beta +sin^2 \beta =1$

$\frac{1+sin \beta }{cos \beta} * \frac{1-sin \beta }{cos \beta} = \frac{(1+sin \beta)(1-sin \beta )}{cos^2 \beta } = \frac{1-sin^2 \beta }{cos^2 \beta } = \frac{cos^2 \beta }{cos^2 \beta } =1$

(sinα+cosα)²-2sinαcosα=sin²α+2sinαcosα+cos²α-2sinαcosα=sin²α+cos²α=1

sin⁴α+cos⁴α+2sin²αcos²α=(sin²α+cos²α)²=1²=1

$\frac{2-sin^2 \alpha -cos^2 \alpha }{3sin^2 \alpha +3cos^2 \alpha } = \frac{2-(sin^2 \alpha +cos^2 \alpha) }{3(sin^2 \alpha +cos^2 \alpha) } = \frac{2-1}{3} = \frac{1}{3}$

$\frac{sin^4 \alpha -cos^4 \alpha }{sin^2 \alpha -cos^2 \alpha} = \frac{(sin^2 \alpha -cos^2 \alpha)(sin^2 \alpha +cos^2 \alpha ) }{sin^2 \alpha -cos^2 \alpha} =sin^2 \alpha +cos^2 \alpha =1$