(C) We know the identity $\sec^{2} \theta = 1 + \tan^{2} \theta$.Substituting the given value: $\sec^{2} \theta = 1 + (\frac{5}{12})^{2}$.$\sec^{2} \t

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) We know the identity $\sec^{2} \theta = 1 + \tan^{2} \theta$.Substituting the given value: $\sec^{2} \theta = 1 + (\frac{5}{12})^{2}$.$\sec^{2} \theta = 1 + \frac{25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$.Taking the square root,$\sec \theta = \frac{13}{12}$.Since $\cos \theta = \frac{1}{\sec \theta}$,we get $\cos \theta = \frac{12}{13}$.