C-09-Q050mediumsingle_mcq1+tan2θ=?1 + \tan^2\theta = ?1+tan2θ=?asec2θ\sec^2\thetasec2θbcosec2θ\operatorname{cosec}^2\thetacosec2θccot2θ\cot^2\thetacot2θdsin2θ\sin^2\thetasin2θব্যাখ্যাDividing the identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1sin2θ+cos2θ=1 by cos2θ\cos^2\thetacos2θ gives tan2θ+1=sec2θ\tan^2\theta + 1 = \sec^2\thetatan2θ+1=sec2θ. Hence 1+tan2θ=sec2θ1 + \tan^2\theta = \sec^2\theta1+tan2θ=sec2θ.