có A=\(\dfrac{1-cosa+2cos^2a-1}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)

\(VT=\dfrac{1+\cos^2a-\sin^2a+2\cdot\sin a\cdot\cos a}{1+2\cdot\sin a\cdot\cos a-\cos^2a+\sin^2a}\)

\(=\dfrac{2\cdot\cos^2a+2\cdot\sin a\cdot\cos a}{2\cdot\sin^2a+2\cdot\sin a\cdot\cos a}\)

\(=\dfrac{2\cdot\cos a\left(\cos a+\sin a\right)}{2\cdot\sin a\cdot\left(\sin a+\cos a\right)}\)

\(=\dfrac{\cos a}{\sin a}=\cot a\)

Gọi tam giác vuông đề bài cho là ΔABC vuông tại A, có \(\hat{B}=\alpha\)

Xét ΔABC vuông tại A có

\(\sin\alpha=\sin B=\frac{AC}{BC};cos\alpha=cosB=\frac{AB}{BC}\)

\(\tan\alpha=\tan B=\frac{AC}{AB};\cot\alpha=\cot B=\frac{AB}{AC}\)

\(\frac{\sin\alpha}{cos\alpha}=\frac{AC}{BC}:\frac{AB}{BC}=\frac{AC}{AB}=tan\alpha\)

\(\frac{cosa}{\sin a}=\frac{cosB}{\sin B}=\frac{AB}{BC}:\frac{AC}{BC}=\frac{AB}{AC}=\cot\alpha\)

\(\tan a\cdot\cot\alpha=\frac{\sin\alpha}{cos\alpha}\cdot\frac{cos\alpha}{\sin\alpha}=1\)

\(\sin^2\alpha+cos^2\alpha=\left(\frac{AB}{BC}\right)^2+\left(\frac{AC}{BC}\right)^2=\frac{AB^2+AC^2}{BC^2}=\frac{BC^2}{BC^2}=1\)

(Sina -cosa)^2 =1:25

<=> sin^2a +cos^2a -2sina.cosa =1:25

Ta có sin^2a+cos^2a = 1 

<=> 1-2 sina.cosa =1:25

2sina.cosa =24:25

CT : sin2a= 2sina.cosa=24:25

 Có sin^2 .2a + co^2.2a = 1 

       (24:25)^2 + cos^2.2a =1 

Từ đây rút cos 2a = căn 1-(24:25)^2 =...  bạn  tự làm tiếp nha !

\(\left(1+tan^2a\right)\left(1-sin^2a\right)-\left(1+cot^2a\right)\left(1-cos^2a\right)\)

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

\(=cos^2a+sin^2a-sin^2a-cos^2a=\)\(0\)

Vậy B=0

\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)

\(=cos^2a+sin^2a+cos^2b+sin^2b+2\left(cosa.cosb+sina.sinb\right)\)

\(=2+2cos\left(a-b\right)=2+2cos\frac{\pi}{3}=3\)

\(\left(cosa+sina\right)^2=\frac{36}{25}\Leftrightarrow1+2sina.cosa=\frac{36}{25}\)

\(\Rightarrow sin2a=\frac{36}{25}-1=\frac{11}{25}\)

\(cos2a=cos^2a-sin^2a=\left(cosa-sina\right)\left(cosa+sina\right)>0\)

\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{6\sqrt{14}}{25}\)

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