12.158. Find the cosines of the acute angles of a right triangle, given that the product of the tangents of half these angles is $1 / 6$.
Solution.

Let one of the angles be $\alpha$. Then the second angle will be $90^{\circ}-\alpha$ and by the condition $\operatorname{tg} \frac{\alpha}{2} \operatorname{tg}\left(45^{\circ}-\frac{\alpha}{2}\right)=\frac{1}{6} \Rightarrow \operatorname{tg} \frac{\alpha}{2} \cdot \frac{1-\operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg} \frac{\alpha}{2}}=\frac{1}{6} \Leftrightarrow$ $\Leftrightarrow 6 \operatorname{tg}^{2} \frac{\alpha}{2}-5 \operatorname{tg} \frac{\alpha}{2}+1=0$;

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0915.jpg?height=383&width=766&top_left_y=77&top_left_x=271)

Fig. 12.31

1) $\operatorname{tg} \frac{\alpha}{2}=\frac{1}{2}$
2) $\operatorname{tg} \frac{\alpha}{2}=\frac{1}{3}$.

Since $\cos \alpha=\frac{1-\operatorname{tg}^{2} \frac{\alpha}{2}}{1+\operatorname{tg} \frac{\alpha}{2}}$, then 1) $\left.\cos \alpha=\frac{3}{5} ; 2\right) \cos \alpha=\frac{4}{5}$ and $\cos \left(90^{\circ}-\alpha\right)=\frac{4}{5}$

or $\cos \left(90^{\circ}-\alpha\right)=\frac{4}{5}$.

Answer: $\frac{3}{5}$ and $\frac{4}{5}$.