\(P=\dfrac{tan\left(-a\right)+2\cdot cota}{3\cdot tan\left(\dfrac{pi}{2}+a\right)}=\dfrac{-tana+2\cdot\dfrac{1}{2}}{3\cdot\left(-cota\right)}\)

\(=\dfrac{-2+1}{3\cdot\dfrac{-1}{2}}=-1:\dfrac{-3}{2}=\dfrac{2}{3}\)

\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)

\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)

\(tan\left(\frac{\pi}{3}-a\right)tan\left(\frac{\pi}{3}+a\right)=\frac{sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)}{cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)}\)

\(=\frac{cos2a-cos\frac{2\pi}{3}}{cos2a+cos\frac{2\pi}{3}}=\frac{cos2a+\frac{1}{2}}{cos2a-\frac{1}{2}}=\frac{2cos2a+1}{2cos2a-1}\)

\(\Rightarrow tana.tan\left(\frac{\pi}{3}-a\right)tan\left(\frac{\pi}{3}+a\right)=\frac{sina\left(2cos2a+1\right)}{cosa\left(2cos2a-1\right)}=\frac{2sina.cos2a+sina}{2cos2a.cosa-cosa}\)

\(=\frac{sin3a-sina+sina}{cos3a+cosa-cosa}=\frac{sin3a}{cos3a}=tan3a\)

\(\left(tana+cota\right)^2=16\)

\(\Leftrightarrow tan^2a+cot^2a+2=16\)

\(\Rightarrow tan^2a+cot^2a=14\)

\(tan^2\left(a+3\pi\right)+tan^2\left(a+\frac{3\pi}{2}\right)=tan^2a+cot^2a=14\)

Có \(a\) thuộc góc phần tư thứ III -> sin\(a\) < 0

+) sin\(a\)=-\(\sqrt{1-cos^2a}\)=-\(\sqrt{1-\left(\dfrac{-12}{13}\right)^2}\)=\(\dfrac{-5}{13}\)

\(cos2a=cos^2a-sin^2a\)=\(\left(\dfrac{-12}{13}\right)^2-\left(\dfrac{-5}{13}\right)^2=\dfrac{119}{169}\)

\(0< a< \frac{\pi}{2}\Rightarrow sina;cosa;tana>0\)

\(tana+\frac{1}{tana}=3\Leftrightarrow tan^2a-3tana+1=0\) \(\Rightarrow\left[{}\begin{matrix}tana=\frac{3-\sqrt{5}}{2}\\tana=\frac{3+\sqrt{5}}{2}\end{matrix}\right.\)

- Với \(tana=\frac{3-\sqrt{5}}{2}\)

\(\Rightarrow cota=\frac{1}{tana}=\frac{3+\sqrt{5}}{2}\)

\(1+tan^2a=\frac{1}{cos^2a}\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}=\frac{2}{\sqrt{18-6\sqrt{5}}}\)

\(sina=\sqrt{1-cos^2a}=\frac{2}{\sqrt{18+6\sqrt{5}}}\)

\(cos\left(\frac{3\pi}{2}-a\right)=cos\left(2\pi-\frac{\pi}{2}-a\right)=-sina=...\)

\(sin\left(2\pi+a\right)=sina=...\)

\(tan\left(\pi-a\right)=-tana=...\)

\(cot\left(\pi+a\right)=cota=...\)

TH2: \(tana=\frac{3+\sqrt{5}}{2}\)

Tương tự như trên

bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)

\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)

\(tanx=tan\alpha\Rightarrow x=\alpha+k\pi\)

Lời giải:

Theo công thức lượng giác:

\(F=\sin (\pi +a)-\cos (\frac{\pi}{2}-a)+\cot (2\pi -a)+\tan (\frac{3\pi}{2}-a)\)

\(=-\sin a-\sin a+\cot (\pi -a)+\tan (\frac{\pi}{2}-a)\)

\(=-2\sin a-\cot a+\cot a=-2\sin a\)