Question

Cos²a+tg²a+sin²a при a=п/6

Answer 1

$\displaystyle\bf\\Cos^{2} \alpha +tg^{2} \alpha +Sin^{2}\alpha = \underbrace{Cos^{2} \alpha +Sin^{2} \alpha}_{1} +tg^{2}\alpha = 1+tg^{2}\alpha \\\\\alpha =\frac{\pi }{6} \\\\1+tg^{2} \alpha =1+tg^{2}\frac{\pi }{6} =1+\Big(\frac{1}{\sqrt{3} }\Big) ^{2} =1+\frac{1}{3} =1\frac{1}{3}$