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

\(=tanx.\dfrac{1+cos^2x-sin^2x}{sinx}\)

\(=\dfrac{sinx}{cosx}.\dfrac{2cos^2x}{sinx}=2cosx\)

\(\Leftrightarrow\left\{{}\begin{matrix}4a+2b+3=-1\\-\dfrac{b}{2a}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+2b=-4\\b=-4a\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-4\end{matrix}\right.\Leftrightarrow a-b=5\)

1.

\(\dfrac{1-cosx+cos2x}{sin2x-sinx}=\dfrac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)

\(=\dfrac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\dfrac{cosx}{sinx}=cotx\)

2.

\(\dfrac{1+tan^4x}{tan^2x+cot^2x}=\dfrac{1+tan^4x}{tan^2x+\dfrac{1}{tan^2x}}=\dfrac{1+tan^4x}{\dfrac{tan^4x+1}{tan^2x}}=tan^2x\)

3.

\(sin^4x+cos^4x=sin^4x+cos^4x+2sin^2x.cos^2x-2sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)

\(=1-2sin^2x.cos^2x\)

4.

Áp dụng câu 3:

\(sin^4x+cos^4x=1-2sin^2x.cos^2x\)

\(=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2\)

\(=1-\dfrac{1}{2}sin^22x\)

5.

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

\(=\dfrac{1}{2}\left(cos2y-cos2x\right)=\dfrac{1}{2}\left(1-2sin^2y\right)-\dfrac{1}{2}\left(1-2sin^2x\right)\)

\(=sin^2x-sin^2y\)

6.

\(tanx+cotx=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}\)

\(=\dfrac{1}{sinx.cosx}=\dfrac{2}{2sinx.cosx}=\dfrac{2}{sin2x}\)

8C

10A

11A

12C

13D

b: \(VT=\left[\dfrac{\dfrac{sinx}{cosx}+sinx}{1+cosx}\right]^2+1\)

\(=\left[\dfrac{sinx\left(\dfrac{1}{cosx}+1\right)}{cosx\left(1+\dfrac{1}{cosx}\right)}\right]^2+1\)

=1/cos^2x=VP