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Answer:[tex]\tan \theta = - \frac{1}{5} = - 0.2[/tex][tex]\cos \theta = 0.98[/tex][tex]\sin \theta = - 0.196[/tex]Step-by-step explanation:It is given that [tex]\cot \theta = - 5[/tex] and [tex]\theta[/tex] is in the fourth quadrant.So, only [tex]\cos \theta[/tex] will have positive value and [tex]\sin \theta[/tex], [tex]\tan \theta[/tex] will have negative value.Now, [tex]\cot \theta = - 5[/tex]⇒ [tex]\tan \theta = \frac{1}{\cot \theta} = -\frac{1}{5}[/tex] (Answer)We know, that [tex]\sec^{2} \theta - \tan^{2} \theta = 1[/tex]⇒ [tex]\sec \theta = \sqrt{1 + \tan^{2} \theta } = \sqrt{1 + (- \frac{1}{5} )^{2} } = 1.019[/tex] {Since, [tex]\cos \theta[/tex] is positive then [tex]\sec \theta[/tex] will also be positive}⇒ [tex]\cos \theta = \frac{1}{\sec \theta} = \frac{1}{1.0198} = 0.98[/tex] (Answer)We know, that [tex]\csc^{2} \theta - \cot^{2} \theta = 1[/tex]⇒ [tex]\csc \theta = - \sqrt{1 + \cot^{2} \theta } = - \sqrt{1 + (- 5 )^{2} } = - 5.099[/tex] {Since, [tex]\sin \theta[/tex] is negative then [tex]\csc \theta[/tex] will also be negative}⇒ [tex]\sin \theta = \frac{1}{\csc \theta} = \frac{1}{- 5.099} = - 0.196[/tex] (Answer)