[tex]\text{Given that,}\\\\\sin \theta = -\dfrac{12}{13}\\\\\implies \sin^2 \theta = \dfrac{144}{169}\\\\\implies 1- \cos^2 \theta = \dfrac{144}{169}\\\\\implies \cos^2 \theta = 1- \dfrac{144}{169}\\\\\implies \cos^2 \theta = \dfrac{25}{169}\\\\\implies \cos \theta = \pm \sqrt{\dfrac{25}{169}} = \pm \dfrac 5{13}\\\\\\\text{Since,}~ \tan \theta >0,\\\\\cos \theta =- \dfrac{5}{13}[/tex]
