C-09-Q170mediumsingle_mcq1−tan260∘1+tan260∘+sin260∘=?\dfrac{1 - \tan^2 60^\circ}{1 + \tan^2 60^\circ} + \sin^2 60^\circ = ?1+tan260∘1−tan260∘+sin260∘=?a14\dfrac{1}{4}41b12\dfrac{1}{2}ব্যাখ্যাWith tan260∘=3\tan^2 60^\circ=3tan260∘=3, the fraction is 1−31+3=−24=−12\frac{1-3}{1+3}=\frac{-2}{4}=-\tfrac12, and . Therefore .