cho `tan ∝=2` tính
a) `\frac{2sin ∝+3cos ∝}{3sin ∝-2cos ∝}`
b) `\frac{2sin^2 ∝+3cos ∝sin ∝+3cos^2 ∝}{4sin ∝+3sin∝cos ∝}`
cho `tan ∝=2` tính
a) `\frac{2sin ∝+3cos ∝}{3sin ∝-2cos ∝}`
b) `\frac{2sin^2 ∝+3cos ∝sin ∝+3cos^2 ∝}{4sin ∝+3sin∝cos ∝}`
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\tan \alpha = 2 \Leftrightarrow \dfrac{{\sin \alpha }}{{\cos \alpha }} = 2 \Rightarrow \cos \alpha \ne 0\\
a,\\
\dfrac{{2\sin \alpha + 3\cos \alpha }}{{3\sin \alpha – 2\cos \alpha }} = \dfrac{{\dfrac{{2\sin \alpha + 3\cos \alpha }}{{\cos \alpha }}}}{{\dfrac{{3\sin \alpha – 2\cos \alpha }}{{\cos \alpha }}}} = \dfrac{{2.\dfrac{{\sin \alpha }}{{\cos \alpha }} + 3}}{{3.\dfrac{{\sin \alpha }}{{\cos \alpha }} – 2}}\\
= \dfrac{{2.\tan \alpha + 3}}{{3.\tan \alpha – 2}} = \dfrac{{2.2 + 3}}{{3.2 – 2}} = \dfrac{7}{4}\\
b,\\
\dfrac{{2{{\sin }^2}\alpha + 3.\cos \alpha .\sin \alpha + 3{{\cos }^2}\alpha }}{{4{{\sin }^2}\alpha + 3.\sin \alpha .\cos \alpha }}\\
= \dfrac{{\dfrac{{2{{\sin }^2}\alpha + 3.\cos \alpha .\sin \alpha + 3{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}}}{{\dfrac{{4{{\sin }^2}\alpha + 3.\sin \alpha .\cos \alpha }}{{{{\cos }^2}\alpha }}}}\\
= \dfrac{{2.\dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + 3.\dfrac{{\sin \alpha }}{{\cos \alpha }} + 3}}{{4.\dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + 3.\dfrac{{\sin \alpha }}{{\cos \alpha }}}}\\
= \dfrac{{2.{{\tan }^2}\alpha + 3.\tan \alpha + 3}}{{4.{{\tan }^2}\alpha + 3.\tan \alpha }}\\
= \dfrac{{{{2.2}^2} + 3.2 + 3}}{{{{4.2}^2} + 3.2}}\\
= \dfrac{{17}}{{22}}
\end{array}\)
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