Question

Помогите ПЖ 9 класс алгебра дам 21 балл нужна зделать 1 и 5 пример​

Answer 1

Объяснение:

$1)\ tg75^0-tg15^0=\frac{sin75^0}{cos75^0} -\frac{sin15^0}{cos15^0}=\frac{sin75^0*cos15^0-sin15^0cos75^0}{cos75^0*cos15^0} =\\=\frac{\frac{1}{2}*(sin(75^0-15^0)+sin(75^0+15^0))-\frac{1}{2}*(sin( 15^0-75^0)+sin(15^0+75^0) )}{\frac{1}{2}*(cos(75^0-15^0)+cos(75^0+15^0)) } =\\\frac{sin60^0+sin90^0-sin(-60^0)-sin90^0}{cos60^0+cos90^0} =\frac{2*\frac{\sqrt{3} }{2} }{\frac{1}{2} +0}=\frac{\sqrt{3} }{\frac{1}{2} }= 2\sqrt{3}.$

$5)\ ctg50^0-ctg20^0=\frac{cos50^0}{sin50^0} -\frac{cos20^0}{sin20^0}=\frac{sin20^0*cos50^0-sin50^0*cos20^0}{sin50^0*sin20^0} =\\=\frac{\frac{1}{2}*(sin(20^0-50^0)+sin(20^0+50^0))-\frac{1}{2} (sin(50^0-20^0)+sin(50^0+20^0)) }{\frac{1}{2}*(cos(50^0-20^0)-cos(50^0+20^0)) } =\\=\frac{sin(-30^0)+sin70^0-sin30^0-sin70^0}{cos30^0-cos70^0} \approx\frac{-2*\frac{1}{2} }{\frac{\sqrt{3} }{2}-0,342 } \approx\frac{-1}{0,866-0,342}\approx-\frac{1}{0,524} \approx-1,91.$