Cho 0 26/07/2021 Bởi Gabriella Cho 0 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " Cho 0
Đáp án: $\begin{array}{l}{\tan ^2}x – {\sin ^2}x\\ = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} – {\sin ^2}x\\ = {\sin ^2}x\left( {\dfrac{1}{{{{\cos }^2}x}} – 1} \right)\\ = {\sin ^2}x.\dfrac{{1 – {{\cos }^2}x}}{{{{\cos }^2}x}}\\ = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}.\left( {1 – {{\cos }^2}x} \right)\\ = {\tan ^2}x.si{n^2}x\\\left( {do:{{\sin }^2}x + {{\cos }^2}x = 1} \right)\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
{\tan ^2}x – {\sin ^2}x\\
= \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} – {\sin ^2}x\\
= {\sin ^2}x\left( {\dfrac{1}{{{{\cos }^2}x}} – 1} \right)\\
= {\sin ^2}x.\dfrac{{1 – {{\cos }^2}x}}{{{{\cos }^2}x}}\\
= \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}.\left( {1 – {{\cos }^2}x} \right)\\
= {\tan ^2}x.si{n^2}x\\
\left( {do:{{\sin }^2}x + {{\cos }^2}x = 1} \right)
\end{array}$