Question

помогите 3 ДАЮ 30пнк

Answer 1

$\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos\alpha=\sqrt{1-\sin^2\alpha}\\\cos\alpha=\sqrt{1-\frac1{64}}=\sqrt{\frac1{36}}=\pm\frac16$
Т.к. угол в I четверти, то 
$\cos\alpha=\frac16\\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac18:\frac16=\frac18\cdot\frac61=\frac68\\ctg\alpha=\frac1{tg\alpha}=1:\frac68=1\cdot\frac86=\frac86$

$\frac5{x-2}-\frac4{x-3}=\frac1x\\\frac{5x-15-4x+8}{(x-2)(x-3)}=\frac1x\\\frac{x-7}{x^2-5x+6}=\frac1x\\x^2-7x=x^2-5x+6\\7x-5x=-6\\2x=-6\\x=-3$