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

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

 

\(5sin2a-6cosa=0\)

\(\Leftrightarrow sin2a=\dfrac{6}{5}cosa\)

\(\Leftrightarrow2\cdot sina\cdot cosa=\dfrac{6}{5}\cdot cosa\)

\(\Leftrightarrow cosa\left(2sina-\dfrac{6}{5}\right)=0\)

=>cosa=0 hoặc sina=3/5

hay \(a=\dfrac{\Pi}{2}+k\Pi\) hoặc \(\left[{}\begin{matrix}a=arcsin\left(\dfrac{3}{5}\right)+k2\Pi\\a=\Pi-arcsin\left(\dfrac{3}{5}\right)+k2\Pi\end{matrix}\right.\)

mà 0<a<pi/2

nên \(a=arcsin\left(\dfrac{3}{5}\right)\)

\(A=sina+sina+cota=2\cdot sina+cota\)

\(=\dfrac{38}{15}\)

30 tháng 7 2019
https://i.imgur.com/qOszLcC.jpg

2tan a-cot a=1

=>2tana-1/tan a=1

=>\(\dfrac{2tan^2a-1}{tana}=1\)

=>2tan^2a-tana-1=0

=>(tan a-1)(2tana+1)=0

=>tan a=-1/2 hoặc tan a=1

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

TH1: tan a=-1/2

\(P=\dfrac{\dfrac{1}{2}+2\cdot\left(-2\right)}{-3\cdot\left(-2\right)}=-\dfrac{7}{2}:6=-\dfrac{7}{12}\)

TH2: tan a=1

=>cot a=1

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

14 tháng 9 2023

Ta có :

\(2tan\alpha-cot\alpha=1\)

\(\Leftrightarrow2tan\alpha-\dfrac{1}{tan\alpha}=1\)

\(\Leftrightarrow2tan\alpha-\dfrac{1}{tan\alpha}-1=0\)

\(\Leftrightarrow\dfrac{2tan^2\alpha-tan\alpha-1}{tan\alpha}=0\left(tan\alpha\ne0\right)\)

\(\Leftrightarrow2tan^2\alpha-tan\alpha-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tan\alpha=1\\tan\alpha=-\dfrac{1}{2}\end{matrix}\right.\)

\(P=\dfrac{tan\left(8\pi-\alpha\right)+2cot\left(\pi+\alpha\right)}{3tan\left(\dfrac{3\pi}{2}+\alpha\right)}\)

\(\Leftrightarrow P=\dfrac{tan\left(4.2\pi-\alpha\right)+2cot\alpha}{3tan\left(2\pi-\dfrac{\pi}{2}+\alpha\right)}\)

\(\Leftrightarrow P=\dfrac{tan\left(-\alpha\right)+2cot\alpha}{3tan\left[-\left(\dfrac{\pi}{2}-\alpha\right)\right]}\)

\(\Leftrightarrow P=\dfrac{-tan\alpha+2cot\alpha}{-3tan\left(\dfrac{\pi}{2}-\alpha\right)}\)

\(\Leftrightarrow P=\dfrac{-tan\alpha+2cot\alpha}{-3cot\alpha}\)

- Với \(tan\alpha=1\Rightarrow cot\alpha=1\)

\(\Leftrightarrow P=\dfrac{-1+2.1}{-3.1}=-\dfrac{1}{3}\)

- Với \(tan\alpha=-\dfrac{1}{2}\Rightarrow cot\alpha=-2\)

\(\Leftrightarrow P=\dfrac{\dfrac{1}{2}+2.\left(-2\right)}{-3.\left(-2\right)}=\dfrac{-\dfrac{7}{2}}{6}=-\dfrac{7}{12}\)

 

NV
19 tháng 10 2019

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

18 tháng 5 2017

a) \(x=-45^0+k90^0,k\in\mathbb{Z}\)

b) \(x=-\dfrac{\pi}{6}+k\pi,k\in\mathbb{Z}\)

c) \(x=\dfrac{3\pi}{4}+k2\pi,k\in\mathbb{Z}\)

d) \(x=300^0+k540^0,k\in\mathbb{Z}\)