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Исбот кунед, ки \(\operatorname{ctg}{70^\circ}+4\cos{70^\circ}=\sqrt{3}\)

\(\operatorname{ctg}{\alpha}=\operatorname{tg}{90^\circ-\alpha}\)

\(\cos{\alpha}=\sin{90^\circ-\alpha}\)

\(\operatorname{ctg}{70^\circ}=\operatorname{tg}{20^\circ}\)

\(\cos{70^\circ}=\sin{20^\circ}\)

\(\operatorname{tg}{20^\circ}+4\sin{20^\circ}=\sqrt{3}\)

\(4\sin{20^\circ}=\sqrt{3}-\operatorname{tg}{20^\circ}\)

\(\sin{20^\circ}=\frac{\sqrt{3}-\operatorname{tg}{20^\circ}}{4}\)

\(\sqrt{3}=\operatorname{tg}{60^\circ}\)

\(\operatorname{tg}{\alpha}-\operatorname{tg}{\beta}=\frac{\sin{(\alpha-\beta)}}{\cos{\alpha}\cdot\cos{\beta}}\)

\(\frac{\sqrt{3}-\operatorname{tg}{20^\circ}}{4}=\frac{\operatorname{tg}{60^\circ}-\operatorname{tg}{20^\circ}}{4}=\)

\(=\frac{\frac{\sin{40^\circ}}{\cos{60^\circ}\cdot\cos{20^\circ}}}{4}=\)

\(=\frac{\sin{40^\circ}}{4\cdot\cos{60^\circ}\cdot\cos{20^\circ}}\)

\(\sin{40^\circ}=2\cdot\sin{20^\circ}\cos{20^\circ}\)

\(\frac{\sin{40^\circ}}{4\cdot\cos{60^\circ}\cdot\cos{20^\circ}}=\)

\(=\frac{2\cdot\sin{20^\circ}\cos{20^\circ}}{4\cdot\cos{60^\circ}\cdot\cos{20^\circ}}\)

\(\cos{60^\circ}=\frac{1}{2}\)

\(\frac{2\cdot\sin{20^\circ}\cos{20^\circ}}{4\cdot\cos{60^\circ}\cdot\cos{20^\circ}}=\)

\(=\frac{2\cdot\sin{20^\circ}\cos{20^\circ}}{4\cdot\frac{1}{2}\cdot\cos{20^\circ}}=\)

\(=\frac{2\cdot\sin{20^\circ}\cos{20^\circ}}{2\cdot\cos{20^\circ}}=\)

\(=\sin{20^\circ}\)

\(\sin{20^\circ}=\sin{20^\circ}\)

Исбот шуд.