Question

\(B=\left(1+\tan^2\right)\left(1-\sin^2\right)-\left(1+cotan^2\right)\left(1-\cos^2\right)\)

Answer 1

ta có : \(B=\left(1+tan^2x\right)\left(1-sin^2x\right)-\left(1+cot^2x\right)\left(1-cos^2x\right)\)

\(=\left(1+\dfrac{sin^2x}{cos^2x}\right)\left(sin^2x+cos^2x-sin^2x\right)-\left(1+\dfrac{cos^2x}{sin^2x}\right)\left(sin^2x+cos^2x-cos^2x\right)\)

\(=\dfrac{sin^2x+cos^2x}{cos^2x}\left(cos^2x\right)-\dfrac{sin^2x+cos^2x}{sin^2x}\left(sin^2x\right)\)

\(=\dfrac{1}{cos^2x}.cos^2x-\dfrac{1}{sin^2x}.\left(sin^2\right)x=1-1=0\)

nhớ ghi góc nha bn :) .