[tex]\displaystyle\bf\\If~tan\,\theta=\frac{4}{3}~~then\\\\sin\,\theta=\frac{4}{\sqrt{4^2+3^2}}=\frac{4}{\sqrt{16+9}}=\frac{4}{\sqrt{25}}=\frac{4}{5}\\\\\boxed{\bf~sin\,\theta=\frac{4}{5}}\\\\\\cos\,\theta=\frac{3}{\sqrt{4^2+3^2}}=\frac{3}{\sqrt{16+9}}=\frac{3}{\sqrt{25}}=\frac{3}{5}\\\\\boxed{\bf~cos\,\theta=\frac{3}{5}}[/tex]
