rút gọn tan x .cos x / sim x +(cos x + sim x ).( cos x -sim x) 29/10/2021 Bởi Clara rút gọn tan x .cos x / sim x +(cos x + sim x ).( cos x -sim x)
$\dfrac{\tan x .\cos x}{\sin x}+(\cos x+\sin x)(\cos x -\sin x)$ $=\dfrac{\sin x}{\sin x}.+\cos^2x-\sin^2x$ $= 1+\cos^2x-\sin^2x$ $= 2\cos^2x$ Bình luận
Đáp án: $\begin{array}{l}\tan x.\dfrac{{{\mathop{\rm cosx}\nolimits} }}{{\sin x}} + \left( {\cos x + \sin x} \right).\left( {{\mathop{\rm cosx}\nolimits} – sinx} \right)\\ = \dfrac{{\sin x}}{{\cos x}}.\dfrac{{\cos x}}{{\sin x}} + {\cos ^2}x – {\sin ^2}x\\ = 1 + {\cos ^2}x – {\sin ^2}x\\ = {\cos ^2}x + {\sin ^2}x + {\cos ^2}x – {\sin ^2}x\\ = 2{\cos ^2}x\end{array}$ Bình luận
$\dfrac{\tan x .\cos x}{\sin x}+(\cos x+\sin x)(\cos x -\sin x)$
$=\dfrac{\sin x}{\sin x}.+\cos^2x-\sin^2x$
$= 1+\cos^2x-\sin^2x$
$= 2\cos^2x$
Đáp án:
$\begin{array}{l}
\tan x.\dfrac{{{\mathop{\rm cosx}\nolimits} }}{{\sin x}} + \left( {\cos x + \sin x} \right).\left( {{\mathop{\rm cosx}\nolimits} – sinx} \right)\\
= \dfrac{{\sin x}}{{\cos x}}.\dfrac{{\cos x}}{{\sin x}} + {\cos ^2}x – {\sin ^2}x\\
= 1 + {\cos ^2}x – {\sin ^2}x\\
= {\cos ^2}x + {\sin ^2}x + {\cos ^2}x – {\sin ^2}x\\
= 2{\cos ^2}x
\end{array}$