Bài tập 6.32 trang 190 SBT Toán 10

a)

\(\begin{array}{l}
\sin 6\alpha \cot 3\alpha  - \cos 6\alpha \\
 = 2\sin 3\alpha \cos 3\alpha .\frac{{\cos 3\alpha }}{{\sin 3\alpha }} - \left( {2{{\cos }^2}3\alpha  - 1} \right)\\
 = 2{\cos ^2}3\alpha  - 2{\cos ^2}3\alpha  + 1 = 1
\end{array}\)

b)

\(\begin{array}{l}
{\left[ {\tan \left( {{{90}^0} - \alpha } \right) - \cot \left( {{{90}^0} + \alpha } \right)} \right]^2} - {\left[ {\cot \left( {{{180}^0} + \alpha } \right) + \cot \left( {{{270}^0} + \alpha } \right)} \right]^2}\\
 = {\left( {\cot \alpha  + \tan \alpha } \right)^2} - {\left( {\cot \alpha  - \tan \alpha } \right)^2}\\
 = {\cot ^2}\alpha  + 2 + {\tan ^2}\alpha  - {\cot ^2}\alpha  + 2 - {\tan ^2}\alpha  = 4
\end{array}\)

c)

\(\begin{array}{l}
\left( {\tan \alpha  - \tan \beta } \right)\cot \left( {\alpha  - \beta } \right) - \tan \alpha \tan \beta \\
 = \frac{{\tan \alpha  - \tan \beta }}{{\tan \left( {\alpha  - \beta } \right)}} - \tan \alpha \tan \beta \\
 = 1 + \tan \alpha \tan \beta  - \tan \alpha \tan \beta  = 1
\end{array}\)

d)

\(\begin{array}{l}
\left( {\cot \frac{\alpha }{3} - \tan \frac{\alpha }{3}} \right).\tan \frac{{2\alpha }}{3}\\
 = \left( {\frac{{\cos \frac{\alpha }{3}}}{{\sin \frac{\alpha }{3}}} - \frac{{\sin \frac{\alpha }{3}}}{{\cos \frac{\alpha }{3}}}} \right).\frac{{\sin \frac{{2\alpha }}{3}}}{{\cos \frac{{2\alpha }}{3}}}\\
 = \frac{{{{\cos }^2}\frac{\alpha }{3} - {{\sin }^2}\frac{\alpha }{3}}}{{\sin \frac{\alpha }{3}\cos \frac{\alpha }{3}}}.\frac{{\sin \frac{{2\alpha }}{3}}}{{\cos \frac{{2\alpha }}{3}}}\\
 = \frac{{\cos \frac{{2\alpha }}{3}}}{{\frac{1}{2}\sin \frac{{2\alpha }}{3}}}.\frac{{\sin \frac{{2\alpha }}{3}}}{{\cos \frac{{2\alpha }}{3}}} = 2
\end{array}\)

-- Mod Toán 10 HỌC247