Question

помогите пж 20 номер пж​

Answer 1

$1)\;\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\cos^4\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2=(1)^2=1\\\\\\2)\;(\sin\alpha+\cos\alpha)^2+(\sin\alpha-\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha+\\+\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha=2\sin^2\alpha+2\cos^2\alpha=2\\\\\\3)\;\sin^4\alpha-\cos^4\alpha+\cos^2\alpha=\left(\sin^2\alpha-\cos^2\alpha\right)\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha=\\=\sin^2\alpha-\cos^2\alpha+\cos^2\alpha=\sin^2\alpha$

$4) \cos^2\alpha+\sin^4\alpha+\sin^2\alpha\cos^2\alpha=\cos^2\alpha+\sin^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\\=\cos^2\alpha+\sin^2\alpha=1$

$5)\;\frac{\cos\alpha\cdot tg\alpha}{\sin^2\alpha}-ctg\alpha\cos\alpha=\frac{\cos\alpha\cdot\frac{\sin\alpha}{\cos\alpha}}{\sin^2\alpha}-\frac{\cos\alpha}{\sin\alpha}\cdot\cos\alpha=\frac1{\sin\alpha}-\frac{\cos^2\alpha}{\sin\alpha}=\frac{1-\cos^2\alpha}{\sin\alpha}=\\\\=\frac{\sin^2\alpha}{\sin\alpha}=\sin\alpha$

$6)\;\sin\alpha-\cos\alpha\left(tg\alpha+ctg\alpha\right)=\sin\alpha-\cos\alpha\left(\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}\right)=\\\\=\sin\alpha-\sin\alpha-\frac{\cos^2\alpha}{\sin\alpha}=-\frac{\cos^2\alpha}{\sin\alpha}$

$7)\;\left(tg\alpha+ctg\alpha\right)^2-\left(tg\alpha-ctg\alpha\right)=tg^2\alpha+2tg\alpha ctg\alpha+ctg^2\alpha-tg^2\alpha+\\\\+2tg\alpha ctg\alpha-ctg^2\alpha=4tg\alpha ctg\alpha=4$

$8)\;\left(\frac1{\sin\alpha}+ctg\alpha\right)\left(\frac1{\sin\alpha}-ctg\alpha\right)=\frac1{\sin^2\alpha}-ctg^2\alpha=\frac1{\sin^2\alpha}-\frac{\cos^2\alpha}{\sin^2\alpha}=\\\\=\frac{1-\cos^2\alpha}{\sin^2\alpha}=\frac{\sin^2\alpha}{\sin^2\alpha}=1$