Question

помогите решить!!!!! плез

Answer 1

$tga=2\\\\a\in (0;\frac{\pi}{2})\; \; \to \; \; \; cosa>0\\\\\boxed {\; 1+tg^2a=\dfrac{1}{cos^2a}\; }\\\\\\cos^2a=\dfrac{1}{1+tg^2a}=\dfrac{1}{1+4}=\dfrac{1}{5}\\\\\\cosa>0\; \; ,\; \; \; \; cosa=+\dfrac{1}{\sqrt5}=\dfrac{\sqrt5}{5}$