A= cos⁴ ∝ + cos² ∝ . sin² ∝ + sin² ∝

=cos⁴ ∝+(cos² ∝+1).sin² ∝

=cos⁴ ∝+(1+cos⁴ ∝)(1-cos⁴ ∝)

=cos⁴ ∝+1-cos⁴ ∝=1

hép ,mi ai giúp đi 

Bạn vt rõ ra nhé 

\(A=\sqrt{sin^2x\left(sin^2x+cos^2x\right)}=\sqrt{sin^2x}\)

=|sinx|

\(A=\frac{2.\left(cosx+1\right)-sin^2x}{sinx.\left(cosx+1\right)}=\frac{2.cosx+2-sin^2x}{sinx.\left(cosx+1\right)}=\frac{2.cosx+1+cos^2x}{sinx.\left(cosx+1\right)}=\frac{\left(cosx+1\right)^2}{sinx.\left(cosx+1\right)}=\frac{cosx+1}{sinx}\)

\(A=\dfrac{sinx+sin3x+sin2x}{cosx+cos3x+cos2x}=\dfrac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\dfrac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=tan2x\)

\(B=\dfrac{1-4\sin^2x\cdot\cos^2x}{\sin^2x+2\sin x\cdot\cos x+\cos^2}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x+4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}=\dfrac{1}{2\sin x\cdot\cos x}\)

Chọn C.

Ta có: 

A = sin( a-16⁰) .cos( a + 14⁰) – sin( a + 14⁰) .cos(a - 16⁰) = sin[ ( a - 17⁰) – (a + 13⁰) ] = sin( -30⁰) = -0,5.

\(A=\dfrac{sin\left(a-b\right)}{cosa.cosb}+\dfrac{sin\left(b-c\right)}{cosb.cosc}+\dfrac{sin\left(c-a\right)}{cosc.cosa}\)

\(=\dfrac{sina.cosb-cosa.sinb}{cosa.cosb}+\dfrac{sinb.cosc-cosb.sinc}{cosb.cosc}+\dfrac{sinc.cosa-cosc.sina}{cosc.cosa}\)

\(=\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}+\dfrac{sinb}{cosb}-\dfrac{sinc}{cosc}+\dfrac{sinc}{cosc}-\dfrac{sina}{cosa}\)

\(=0\)