Given:-
[tex]\sin \theta=\frac{12}{13}[/tex]
To find:-
The value of cos theta.
We know the formula,
[tex]\sin ^2\theta+\cos ^2\theta=1[/tex]
From this formula, we derive the value of cos theta. we get,
[tex]\begin{gathered} \sin ^2\theta+\cos ^2\theta=1 \\ \cos \theta=\sqrt[]{1-\sin ^2\theta} \end{gathered}[/tex]
Now we substitute the value of sin theta. so we get,
[tex]\begin{gathered} \cos \theta=\sqrt[]{1-\frac{12^2}{13^2}} \\ \cos \theta=\sqrt[]{1-\frac{144}{169}} \\ \cos \theta=\sqrt[]{\frac{169-144}{169}} \\ \cos \theta=\sqrt[]{\frac{25}{169}} \\ \cos \theta=\frac{5}{13} \end{gathered}[/tex]
So the value of cos theta is,
[tex]\frac{5}{13}[/tex]
