C-09-Q056mediumsingle_mcq1−sin2θ=?1 - \sin^2\theta = ?1−sin2θ=?acos2θ\cos^2\thetacos2θbtan2θ\tan^2\thetatan2θccot2θ\cot^2\thetacot2θdsec2θ\sec^2\thetasec2θব্যাখ্যাFrom the identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1sin2θ+cos2θ=1, subtracting sin2θ\sin^2\thetasin2θ from both sides leaves cos2θ=1−sin2θ\cos^2\theta = 1 - \sin^2\thetacos2θ=1−sin2θ.