\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)

\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)

\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)

\(\cos \alpha  =  - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}}  =  - \frac{{12}}{{13}}\) (vì \(\pi  < \alpha  < \frac{{3\pi }}{2}\))

\(\sin \left( {\alpha  + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)

\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha  + \sin \frac{\pi }{4}sin\alpha  = \frac{{ - 17\sqrt 2 }}{{26}}\)

--.--  \(-\pi>-\frac{3}{2}\pi\) mà
Chắc nhầm đề rồi, phải là \(-\pi>a>-\frac{3}{2}\pi\)mới đúng chứ

\(-\pi>a>-\frac{3}{2}\pi\Leftrightarrow\pi>a>\frac{1}{2}\pi\)

\(\cos a=-\frac{4}{5}\Rightarrow\sin a=\frac{3}{5}\)

\(\sin2a=2\sin a.\cos a=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)

\(\cos2a=2\cos^2a-1=\frac{7}{25}\)

\(\sin\left(\frac{5\pi}{2}-a\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a=-\frac{4}{5}\)

\(\sin\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{3}{5}-\frac{4}{5}.\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}\)

\(\cos\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{-4}{5}-\frac{\sqrt{2}}{2}.\frac{3}{5}=-\frac{7\sqrt{2}}{10}\)

\(\Rightarrow\tan\left(a+\frac{\pi}{4}\right)=\frac{1}{7}\)

\(\cos^2\left(\frac{a}{2}\right)=\frac{1+\cos a}{2}=\frac{1}{10}\Leftrightarrow\left|\cos\frac{a}{2}\right|=\frac{\sqrt{10}}{10}\)

Mà \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)

\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)

\(a\in\left(\frac{\pi}{2};\pi\right)\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{4}{5}\)

\(A=\frac{sin\left(4\pi-\frac{\pi}{2}-a\right)}{sin\left(a+\frac{\pi}{4}\right)-cosa}=\frac{-sin\left(a+\frac{\pi}{2}\right)}{sin\left(a+\frac{\pi}{4}\right)-cosa}=\frac{-cosa}{sina.cos\frac{\pi}{4}+cosa.sin\frac{\pi}{4}-cosa}\)

\(=\frac{-\frac{4}{5}}{\frac{3}{5}.\frac{\sqrt{2}}{2}-\frac{4}{5}.\frac{\sqrt{2}}{2}-\frac{4}{5}}=...\)

\(\frac{3\pi}{2}< a< 2\pi\Rightarrow cosa>0\Rightarrow cosa=\sqrt{1-sin^2a}=\frac{4}{5}\)

\(tana=\frac{sina}{cosa}=-\frac{3}{4}\)

\(sin2a=2sina.cosa=-\frac{24}{25}\)

\(cos2a=2cos^2a-1=\frac{7}{25}\)

\(tan\left(a+\frac{\pi}{4}\right)=\frac{tana+tan\frac{\pi}{4}}{1-tana.tan\frac{\pi}{4}}=\frac{-\frac{3}{4}+1}{1+\frac{3}{4}}=...\)

c sai đề

\(sin\left(a+\frac{\pi}{4}\right)=sina.cos\frac{\pi}{4}+cosa.sin\frac{\pi}{4}=...\)

\(M=\frac{\left(-\frac{3}{5}\right)^2-\left(\frac{7}{25}\right)^2}{-\frac{3}{4}}=...\)

Ta có:

a) \(\sin \left( {\alpha  + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3  + 3\sqrt 2 }}{6}\)      

b) \(\cos \left( {\alpha  + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} =  - \frac{{3 + \sqrt 6 }}{6}\)

c) \(\sin \left( {\alpha  - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)

d) \(\cos \left( {\alpha  - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)

Câu 1:

\(tan\left(a+\frac{\pi}{4}\right)=1\Rightarrow a+\frac{\pi}{4}=\frac{\pi}{4}+k\pi\Rightarrow a=k\pi\) (\(k\in Z\) )

Do \(\frac{\pi}{2}< a< 2\pi\Rightarrow\frac{\pi}{2}< k\pi< 2\pi\Rightarrow\frac{1}{2}< k< 2\Rightarrow k=1\Rightarrow a=\pi\)

\(\Rightarrow P=cos\left(\pi-\frac{\pi}{6}\right)+sin\pi=-\frac{\sqrt{3}}{2}\)

Câu 2:

\(cot\left(a+\frac{\pi}{3}\right)=-\sqrt{3}=cot\left(-\frac{\pi}{6}\right)\)

\(\Rightarrow a+\frac{\pi}{3}=-\frac{\pi}{6}+k\pi\Rightarrow a=-\frac{\pi}{2}+k\pi\) (\(k\in Z\))

\(\Rightarrow\frac{\pi}{2}< -\frac{\pi}{2}+k\pi< 2\pi\Rightarrow-\pi< k\pi< \frac{5\pi}{2}\)

\(\Rightarrow-1< k< \frac{5}{2}\Rightarrow k=\left\{0;1;2\right\}\Rightarrow a=\left\{-\frac{\pi}{2};\frac{\pi}{2};\frac{3\pi}{2}\right\}\) \(\Rightarrow cosa=0\)

\(\Rightarrow P=sin\left(\pi+\frac{\pi}{6}\right)+0=-sin\frac{\pi}{6}=-\frac{1}{2}\)

Câu 1:

\(tan\left(a+\frac{\pi}{4}\right)=1\Leftrightarrow\frac{sin\left(a+\frac{\pi}{4}\right)}{cos\left(a+\frac{\pi}{4}\right)}=1\Leftrightarrow sin\left(a+\frac{\pi}{4}\right)=cos\left(a+\frac{\pi}{4}\right)\)

\(\Leftrightarrow sina.cos\frac{\pi}{4}+cosa.sin\frac{\pi}{4}=cosa.cos\frac{\pi}{4}-sina.sin\frac{\pi}{4}\)

\(\Leftrightarrow\frac{\sqrt{2}}{2}sina+\frac{\sqrt{2}}{2}cosa=\frac{\sqrt{2}}{2}cosa-\frac{\sqrt{2}}{2}sina\)

\(\Rightarrow\sqrt{2}sina=0\Rightarrow sina=0\)

\(\Rightarrow P=cos\left(\pi-\frac{\pi}{6}\right)+0=-cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}\)

Câu 2:

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

\(\Leftrightarrow cos\left(a+\frac{\pi}{3}\right)+\sqrt{3}sin\left(a+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow cosa.cos\frac{\pi}{3}-sina.sin\frac{\pi}{3}+\sqrt{3}sina.cos\frac{\pi}{3}+\sqrt{3}cosa.sin\frac{\pi}{3}=0\)

\(\Leftrightarrow\frac{1}{2}cosa-\frac{\sqrt{3}}{2}sina+\frac{\sqrt{3}}{2}sina+\frac{3}{2}cosa=0\)

\(\Leftrightarrow2cosa=0\Rightarrow cosa=0\)

\(\Rightarrow P=sin\left(\pi+\frac{\pi}{6}\right)+0=-sin\frac{\pi}{6}=-\frac{1}{2}\)

Này là kiến thức lớp 10 mà bạn...

\(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\)

\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{4}{5}\)

\(P=1-\left[1-cos\left(\frac{\pi}{2}-2a\right)\right]+sin2a-cos2a-6cota\)

\(=sin2a+sin2a-cos2a-6cota\)

\(=2sin2a-cos2a-6cota\)

\(=4sina.cosa-\left(cos^2a-sin^2a\right)-\frac{6cosa}{sina}\) (thay số và bấm máy)