Question

If tan(t)=3/4 what’s cos(t)

Answer 1

well, let's notice our tangent is positive, that only happens on the 1st and 3rd Quadrants

`tan(t )=\cfrac{\stackrel{opposite}{3}}{\underset{adjacent}{4}}\hspace{5em} \textit{now let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=√(a^2 + b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√(4^2 + 3^2)\implies c=5\hspace{5em} \stackrel{I~Quadrant}{cos(t)= \cfrac{\stackrel{adjacent}{4}}{\underset{hypotenuse}{5}}}\qquad \stackrel{III~Quadrant}{cos(t)=-\cfrac{4}{5}}`