C-09-Q174mediumsingle_mcqcos230∘−sin230∘=?\cos^2 30^\circ - \sin^2 30^\circ = ?cos230∘−sin230∘=?acos60∘\cos 60^\circcos60∘bsin60∘\sin 60^\circsin60∘ccos90∘\cos 90^\circcos90∘d000ব্যাখ্যাThis matches the identity cos2θ−sin2θ=cos2θ\cos^2\theta-\sin^2\theta=\cos 2\thetacos2θ−sin2θ=cos2θ with θ=30∘\theta=30^\circθ=30∘, giving cos60∘\cos 60^\circcos60∘. Numerically, 34−14=12=cos60∘\tfrac34-\tfrac14=\tfrac12=\cos 60^\circ43−41=21=cos60∘.