Добрый день, помогите с решением тригонометрии! Кто сколько может. с 34 по 37 номер. Заранее спасибо. (решение фоткой, пожалуйста)

34. [latex]\frac{2cos40^\circ-sin70^\circ}{sin340^\circ}=\frac{2cos(60^\circ-20^\circ)-sin(90^\circ-20^\circ)}{sin(360^\circ-20^\circ)}=\\\\ =\frac{2(cos60^\circ\cdot cos20^\circ-sin60^\circ\cdot sin20^\circ)-cos20^\circ}{-sin20^\circ}=\\\\ =\frac{cos20^\circ-2(\frac{1}{2}\cdot cos20^\circ-\frac{\sqrt{3}}{2}\cdot sin20^\circ)}{sin20^\circ}=\\\\ =\frac{cos20^\circ-cos20^\circ+\sqrt{3}\cdot sin20^\circ)}{sin20^\circ}=\\\\ =\sqrt{3}[/latex] Ответ: 1) [latex]\sqrt3[/latex]. 35. [latex]\frac{2sin170^\circ+cos40^\circ}{cos130^\circ}=\frac{2sin(180^\circ-10^\circ)+cos(30^\circ+10^\circ)}{cos(90^\circ+40^\circ)}=\\\\ =\frac{2sin10^\circ+cos(30^\circ+10^\circ)}{-sin40^\circ}=\\\\ =-\frac{2sin10^\circ+cos30^\circ\cdot cos10^\circ-sin30^\circ\cdot sin10^\circ}{sin30^\circ\cdot cos10^\circ+sin10^\circ\cdot cos30^\circ}=\\\\ =-\frac{2sin10^\circ+\frac{\sqrt3}{2}\cdot cos10^\circ-\frac{1}{2}\cdot sin10^\circ}{sin30^\circ\cdot cos10^\circ+sin10^\circ\cdot cos30^\circ}=[/latex] [latex]=-\frac{\frac{3}{2}sin10^\circ+\frac{\sqrt3}{2}\cdot cos10^\circ}{\frac{1}{2}\cdot cos10^\circ+\frac{\sqrt3}{2}\cdot sin10^\circ}=\\\\ =-\frac{3sin10^\circ+\sqrt3\cdot cos10^\circ}{\sqrt3\cdot sin10^\circ+cos10^\circ}=\\\\ =-\sqrt3\cdot \frac{\sqrt3\cdot sin10^\circ+cos10^\circ}{\sqrt3\cdot sin10^\circ+cos10^\circ}=\\\\ =-\sqrt3 [/latex] Ответ: 1) [latex]-\sqrt3[/latex]. 36. [latex]\frac{1}{cos105^\circ}+\frac{\sqrt3}{sin285^\circ}=\frac{1}{cos(90^\circ+15^\circ)}+\frac{\sqrt3}{sin(360^\circ-75^\circ)}=\\\\ =-\frac{1}{sin15^\circ}-\frac{\sqrt3}{sin75^\circ}=\\\\ =-\frac{1}{sin15^\circ}-\frac{\sqrt3}{sin(90^\circ-15^\circ)}=\\\\ =-\frac{1}{sin15^\circ}-\frac{\sqrt3}{cos15^\circ}=\\\\ =-\frac{cos15^\circ+\sqrt3\cdot sin15^\circ}{sin15^\circ\cdot cos15^\circ}=\\\\ =-2\cdot\frac{cos15^\circ+\sqrt3\cdot sin15^\circ}{2\cdot sin15^\circ\cdot cos15^\circ} [/latex] [latex]=-2\cdot\frac{cos15^\circ+\sqrt3\cdot sin15^\circ}{sin30^\circ}=\\\\ =-4\cdot(cos15^\circ+\sqrt3\cdot sin15^\circ)=\\\\ =-8\cdot(\frac{1}{2}\cdot cos15^\circ+\frac{\sqrt3}{2}\cdot sin15^\circ)=\\\\ =-8\cdot(sin30^\circ\cdot cos15^\circ+cos30^\circ\cdot sin15^\circ)=\\\\ =-8\cdot(sin(30^\circ+15^\circ)=-8\cdot sin45^\circ=\\\\ =-4\sqrt2[/latex] Ответ: 2) [latex]-4\sqrt2[/latex]. 37. [latex]\frac{\sqrt2\cdot sin\frac{\pi}{8}-2\cdot cos\frac{5\pi}{8}}{sin\frac{5\pi}{8}}=\frac{\sqrt2\cdot sin\frac{\pi}{8}-2\cdot cos(\frac{\pi}{2}+\frac{\pi}{8})}{sin(\frac{\pi}{2}+\frac{\pi}{8})}= \\\\ =\frac{\sqrt2\cdot sin\frac{\pi}{8}+2\cdot sin\frac{\pi}{8}}{cos\frac{\pi}{8}}= \\\\ =(2+\sqrt2)\cdot \frac{sin\frac{\pi}{8}}{cos\frac{\pi}{8}}=(2+\sqrt2)\cdot tg\frac{\pi}{8} \\\\\\ tg\frac{\pi}{8}=\sqrt{\frac{1-cos\frac{\pi}{4}}{1+cos\frac{\pi}{4}}}= \sqrt{\frac{1-\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}}=[/latex] [latex]=\sqrt{\frac{2-\sqrt2}{2+\sqrt2}}=\sqrt2-1 \\\\\\ (2+\sqrt2)\cdot(\sqrt2-1)=\sqrt2[/latex] Ответ: 5) [latex]\sqrt2[/latex]. Если тех не читается, то во вложениях фото.