Bài tập 34 trang 207 SGK Toán 10 NC
Bài tập 34 trang 207 SGK Toán 10 NC
Chứng minh rằng:
a) \(\frac{{1 - 2\sin \alpha \cos \alpha }}{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }} = \frac{{1 - \tan \alpha }}{{1 + \tan \alpha }}\)
b) \({\tan ^2}\alpha - {\sin ^2}\alpha = {\tan ^2}\alpha .{\sin ^2}\)
c) \(2\left( {1 - \sin \alpha } \right)\left( {1 + \cos \alpha } \right) = {\rm{ }}{\left( {1 - \sin \alpha + \cos \alpha } \right)^2}\)
a)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{{1 - 2\sin \alpha \cos \alpha }}{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}\\
= \frac{{{{\left( {\cos \alpha - \sin \alpha } \right)}^2}}}{{\left( {\cos \alpha - \sin \alpha } \right)\left( {\cos \alpha + \sin \alpha } \right)}}
\end{array}\\
\begin{array}{l}
= \frac{{\cos \alpha - \sin \alpha }}{{\cos \alpha + \sin \alpha }}\\
= \frac{{\cos \alpha \left( {1 - \tan \alpha } \right)}}{{\cos \alpha \left( {1 + \tan \alpha } \right)}} = \frac{{1 - \tan \alpha }}{{1 + \tan \alpha }}
\end{array}
\end{array}\)
b)
\(\begin{array}{l}
{\tan ^2}\alpha - {\sin ^2}\alpha \\
= {\tan ^2}\alpha \left( {1 - {{\cos }^2}\alpha } \right)\\
= {\tan ^2}\alpha .{\sin ^2}\alpha
\end{array}\)
c)
\(\begin{array}{*{20}{l}}
{2\left( {1 - \sin \alpha } \right)\left( {1 + \cos \alpha } \right)}\\
{ = 2 - 2\sin \alpha + 2\cos \alpha - 2\sin \alpha \cos \alpha }\\
\begin{array}{l}
= 1 + {\sin ^2}\alpha + {\cos ^2}\alpha - 2\sin \alpha \\
\,\,\,\,\, + 2\cos \alpha - 2\sin \alpha \cos \alpha
\end{array}\\
{ = {{\left( {1 - \sin \alpha + \cos \alpha } \right)}^2}}
\end{array}\)
-- Mod Toán 10
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