cho \(\Delta ABC\) vuông tại A (hình tự vẽ)

đặt \(\alpha\)tại <B

Ta có: sin \(\alpha\)=\(\dfrac{AC}{BC}\)

tan \(\alpha\)=\(\dfrac{AC}{AB}\)

=> sin \(\alpha\) < tan \(\alpha\)(đpcm)

cos \(\alpha\)= \(\dfrac{AB}{BC}\)

cot \(\alpha\)= \(\dfrac{AB}{AC}\)

=> cos \(\alpha\)< cot \(\alpha\)(đpcm)

\(A+B+C=\pi\Rightarrow\dfrac{A}{2}+\dfrac{B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}\)

\(\Rightarrow tan\left(\dfrac{A}{2}+\dfrac{B}{2}\right)=tan\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)\)

\(\Rightarrow\dfrac{tan\dfrac{A}{2}+tan\dfrac{B}{2}}{1-tan\dfrac{A}{2}tan\dfrac{B}{2}}=cot\dfrac{C}{2}=\dfrac{1}{tan\dfrac{C}{2}}\)

\(\Rightarrow\left(tan\dfrac{A}{2}+tan\dfrac{B}{2}\right)tan\dfrac{C}{2}=1-tan\dfrac{A}{2}tan\dfrac{B}{2}\)

\(\Rightarrow tan\dfrac{A}{2}tan\dfrac{B}{2}+tan\dfrac{B}{2}tan\dfrac{C}{2}+tan\dfrac{C}{2}tan\dfrac{A}{2}=1\)

\(A+B+C=180^0\Rightarrow tan\left(A+B\right)=-tanC\)

\(\Rightarrow\frac{tanA+tanB}{1-tanA.tanB}=-tanC\Leftrightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)

\(\Leftrightarrow tanA+tanB+tanC=tanA.tanB.tanC\)

\(2A+2B+2C=360^0\Rightarrow tan\left(2A+2B\right)=-tan2C\)

\(\Leftrightarrow\frac{tan2A+tan2B}{1-tan2A.tan2B}=-tan2C\)

\(\Leftrightarrow tan2A+tan2B+tan2C=tan2A.tan2B.tan2C\)

ta có: A\2+B\2 = π\2 - C\2 

⇒ tan(A\2+B\2) = tan(π\2 -C\2) 

⇒ (tanA\2 +tanB\2)\[1 - tanA\2.tanB\2] = cotgC\2 

⇒ (tanA\2 +tanB\2).tanC\2 = [1 - tanA\2.tanB\2] 

⇒ tanA\2.tanB\2 + tanB\2.tanC\2 + tanC\2.tanA\2 = 1 

............đpcm............

Ta có:

 \(\begin{array}{l}2a = \left( {a + b} \right) + \left( {a - b} \right) \Rightarrow \tan 2a = \tan \left[ {\left( {a + b} \right) + \left( {a - b} \right)} \right]\\2b = \left( {a + b} \right) - \left( {a - b} \right) \Rightarrow \tan 2b = \tan \left[ {\left( {a + b} \right) - \left( {a - b} \right)} \right]\end{array}\)

\(\begin{array}{l}\tan \left[ {\left( {a + b} \right) + \left( {a - b} \right)} \right] = \frac{{\tan \left( {a + b} \right) + \tan \left( {a - b} \right)}}{{1 - \tan \left( {a + b} \right).\tan \left( {a - b} \right)}} = \frac{{3 + 2}}{{1 - 3.2}} =  - 1\\\tan \left[ {\left( {a + b} \right) - \left( {a - b} \right)} \right] = \frac{{\tan \left( {a + b} \right) - \tan \left( {a - b} \right)}}{{1 + \tan \left( {a + b} \right).\tan \left( {a - b} \right)}} = \frac{{3 - 2}}{{1 + 3.2}} = \frac{1}{7}\end{array}\)

Vậy \(\tan 2a =  - 1,\,\,\,\tan 2b = \frac{1}{7}\)

a: sin a=2/3

=>cos^2a=1-(2/3)^2=5/9

=>\(cosa=\dfrac{\sqrt{5}}{3}\)

\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)

\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)

b: cos a=1/5

=>sin^2a=1-(1/5)^2=24/25

=>\(sina=\dfrac{2\sqrt{6}}{5}\)

\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)

\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)

c: cot a=1/tana=1/2

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>1/cos^2a=1+4=5

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)