Đề đúng: \(cos^2\alpha-cos^2\beta=sin^2\beta-sin^2\alpha=\dfrac{1}{1+tan^2\alpha}-\dfrac{1}{1+tan^2\beta}\)

Áp dụng công thức: \(sin^2x+cos^2x=1\Rightarrow cos^2x=1-sin^2x\)

Ta có:

\(cos^2\alpha-cos^2\beta=\left(1-sin^2\alpha\right)-\left(1-sin^2\beta\right)=-sin^2\alpha+sin^2\beta=sin^2\beta-sin^2\alpha\) (1)

Lại có:

\(cos^2\alpha-cos^2\beta=\dfrac{cos^2\alpha}{1}-\dfrac{cos^2\beta}{1}=\dfrac{cos^2\alpha}{sin^2\alpha+cos^2\alpha}-\dfrac{cos^2\beta}{sin^2\beta+cos^2\beta}\)

\(=\dfrac{\dfrac{cos^2\alpha}{cos^2\alpha}}{\dfrac{sin^2\alpha}{cos^2\alpha}+\dfrac{cos^2\alpha}{cos^2\alpha}}-\dfrac{\dfrac{cos^2\beta}{cos^2\beta}}{\dfrac{sin^2\beta}{cos^2\beta}+\dfrac{cos^2\beta}{cos^2\beta}}=\dfrac{1}{tan^2\alpha+1}-\dfrac{1}{tan^2\beta+1}\) (2)

(1);(2) suy ra đpcm

Lời giải:

Ta có:

\(A=4\sin ^4a\cos ^2a+(\sin ^2a-\cos ^2a)^2+4\cos ^4a\sin ^2a\)

\(=4\sin ^2a\cos ^2a(\sin ^2a+\cos ^2a)+(\sin ^2a-\cos ^2a)^2\)

\(=4\sin ^2a\cos ^2a+(\sin ^2a-\cos ^2a)^2\)

\(=4\sin ^2a\cos ^2a+\sin ^4a+\cos ^4a-2\sin ^2a\cos ^2a\)

\(=2\sin ^2a\cos ^2a+\sin ^4a+\cos ^4a=(\sin ^2a+\cos ^2a)^2\)

\(=1^2=1\)

Vậy biểu thức có giá trị không phụ thuộc vào $a$

\(90^0< a< 180^0\Rightarrow cosa< 0\)

\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{5}}{3}\)

\(sin2a=2sina.cosa=-\frac{4\sqrt{5}}{9}\)

\(sin\left(a+30^0\right)=sina.cos30^0+cosa.sin30^0=\frac{2}{3}.\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{3}.\frac{1}{2}=\frac{\sqrt{3}}{3}-\frac{\sqrt{5}}{6}\)

Ta thấy : \(\hept{\begin{cases}AD\perp DC\\MP\perp AD\end{cases}}\) \(\Rightarrow PM//DC\)

\(\Rightarrow\frac{MP}{CD}=\frac{AM}{AC}\) ( định lý Talet )

Chứng minh tương tự ta có : \(MN//AB\)

\(\Rightarrow\frac{MN}{AB}=\frac{MC}{AC}\) ( định lý Talet )

Khi đó : \(\frac{MN}{AB}+\frac{MP}{CD}=\frac{AM}{AC}+\frac{MC}{AC}=\frac{AC}{AC}=1\) (ĐPCM)

=tan²\(a\).( cos²\(a\)+ cos²\(a\) + sin²\(a\) - 1)

=tan²\(a\)( cos²\(a\)-1)