1. Cho cos x= -4/5. Tính A= tan x + cot x/ tan x – cot x
2. Cho cot x = -2. Tính B= 2/ $sin^{2}$ x -3sinxcosx + 2 $cos^{2}$ x
Mn giúp mình nha, thank you ^^
1. Cho cos x= -4/5. Tính A= tan x + cot x/ tan x – cot x
2. Cho cot x = -2. Tính B= 2/ $sin^{2}$ x -3sinxcosx + 2 $cos^{2}$ x
Mn giúp mình nha, thank you ^^
Giải thích các bước giải:
$\begin{array}{l}
1)A = \dfrac{{\tan x + \cot x}}{{\tan x – \cot x}}\\
= \dfrac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x – 1}}\\
= \dfrac{{\dfrac{1}{{{{\cos }^2}x}}}}{{\dfrac{1}{{{{\cos }^2}x}} – 2}}\\
= \dfrac{1}{{1 – 2{{\cos }^2}x}}\\
= \dfrac{1}{{1 – 2.{{\left( {\dfrac{{ – 4}}{5}} \right)}^2}}}\\
= \dfrac{{ – 25}}{7}\\
2)B = \dfrac{2}{{{{\sin }^2}x – 3\sin x\cos x + 2{{\cos }^2}x}}\\
= \dfrac{{2\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^2}x – 3\sin x\cos x + 2{{\cos }^2}x}}\\
= \dfrac{{2\left( {1 + {{\left( {\dfrac{{\cos x}}{{\sin x}}} \right)}^2}} \right)}}{{1 – 3.\dfrac{{\cos x}}{{\sin x}} + 2{{\left( {\dfrac{{\cos x}}{{\sin x}}} \right)}^2}}}\\
= \dfrac{{2\left( {1 + {{\cot }^2}x} \right)}}{{1 – 3\cot x + 2{{\cot }^2}x}}\\
= \dfrac{{2\left( {1 + {{\left( { – 2} \right)}^2}} \right)}}{{1 – 3.\left( { – 2} \right) + {{\left( { – 2} \right)}^2}}}\\
= \dfrac{{10}}{{11}}
\end{array}$