Đơn giản các biểu thức a,A=cos^2a+cos^2a.cot^a b,B=sin^x+sin^2x.tan^2x 13/10/2021 Bởi Liliana Đơn giản các biểu thức a,A=cos^2a+cos^2a.cot^a b,B=sin^x+sin^2x.tan^2x
a, $A= \cos^2a+cos^2a.cot^2a$ $=\dfrac{\sin^2a.\cos^2a+\cos^4a}{\sin^2a}$ $= \dfrac{\cos^2a(\sin^2a+\cos^2a}{\sin^2a}$ $=\dfrac{\cos^2a}{\sin^2a}$ $=\cot^2a$ b, (tương tự) Bình luận
Giải thích các bước giải: Ta có: \(\begin{array}{l}a,\\A = {\cos ^2}a + {\cos ^2}a.{\cot ^2}a = {\cos ^2}a.\left( {1 + {{\cot }^2}a} \right)\\ = {\cos ^2}a.\left( {1 + \dfrac{{{{\cos }^2}a}}{{{{\sin }^2}a}}} \right) = {\cos ^2}a.\dfrac{{{{\sin }^2}a + {{\cos }^2}a}}{{{{\sin }^2}a}}\\ = {\cos ^2}a.\dfrac{1}{{{{\sin }^2}a}} = \dfrac{{{{\cos }^2}a}}{{{{\sin }^2}a}} = {\cot ^2}a\\B = {\sin ^2}x + {\sin ^2}x.{\tan ^2}x = {\sin ^2}x\left( {1 + {{\tan }^2}x} \right)\\ = {\sin ^2}x.\left( {1 + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right) = {\sin ^2}x.\dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}}\\ = {\sin ^2}x.\dfrac{1}{{{{\cos }^2}x}} = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\tan ^2}x\end{array}\) Bình luận
a,
$A= \cos^2a+cos^2a.cot^2a$
$=\dfrac{\sin^2a.\cos^2a+\cos^4a}{\sin^2a}$
$= \dfrac{\cos^2a(\sin^2a+\cos^2a}{\sin^2a}$
$=\dfrac{\cos^2a}{\sin^2a}$
$=\cot^2a$
b, (tương tự)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = {\cos ^2}a + {\cos ^2}a.{\cot ^2}a = {\cos ^2}a.\left( {1 + {{\cot }^2}a} \right)\\
= {\cos ^2}a.\left( {1 + \dfrac{{{{\cos }^2}a}}{{{{\sin }^2}a}}} \right) = {\cos ^2}a.\dfrac{{{{\sin }^2}a + {{\cos }^2}a}}{{{{\sin }^2}a}}\\
= {\cos ^2}a.\dfrac{1}{{{{\sin }^2}a}} = \dfrac{{{{\cos }^2}a}}{{{{\sin }^2}a}} = {\cot ^2}a\\
B = {\sin ^2}x + {\sin ^2}x.{\tan ^2}x = {\sin ^2}x\left( {1 + {{\tan }^2}x} \right)\\
= {\sin ^2}x.\left( {1 + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} \right) = {\sin ^2}x.\dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}}\\
= {\sin ^2}x.\dfrac{1}{{{{\cos }^2}x}} = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = {\tan ^2}x
\end{array}\)