= 1

\(P=cos\left(180^o-\alpha\right).tan\alpha+sin\left(180-\alpha\right)\\ =-cos\alpha.\dfrac{sin\alpha}{cos\alpha}+sin\alpha\\ =-sin\alpha+sin\alpha=0\)

=> Chọn A

A= cos⁴ ∝ + cos² ∝ . sin² ∝ + sin² ∝

=cos⁴ ∝+(cos² ∝+1).sin² ∝

=cos⁴ ∝+(1+cos⁴ ∝)(1-cos⁴ ∝)

=cos⁴ ∝+1-cos⁴ ∝=1

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

=|sinx|

hép ,mi ai giúp đi 

Bạn vt rõ ra nhé 

Chọn C.

Ta có:

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.

Chọn C.

Ta có A =  sin ²a.tan²a + 4sin²a - tan²a + 3cos²a

= 3sin²a + 3cos²a = 3.

\(A=\dfrac{sin2\alpha+sin5\alpha-sin3\alpha}{1+cos\alpha-2sin^22\alpha}\)

\(=\dfrac{2sin\alpha.cos\alpha+2.cos4\alpha.sin\alpha}{cos4\alpha+cos\alpha}\)

\(=\dfrac{2sin\alpha.\left(cos\alpha+cos4\alpha\right)}{cos4\alpha+cos\alpha}=2sin\alpha\)