Question

Вычислите:
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Answer 1

Ответ:

1.

$\frac{ \sin( \beta ) \cos( \beta ) + 2}{5 \cos {}^{2} ( \beta ) + 1} = \\ = \frac{ \sin( \beta ) \cos( \beta ) + 2 \sin {}^{2} ( \beta ) + 2\cos {}^{2} ( \beta ) }{5 \cos {}^{2} ( \beta ) + \sin {}^{2} ( \beta ) + \cos {}^{2} ( \beta ) } = \\ = \frac{ \sin( \beta ) \cos( \beta ) + 2 \sin {}^{2} ( \beta ) + 2\cos {}^{2} ( \beta ) }{6 \cos {}^{2} ( \beta ) + \sin {}^{2} ( \beta ) } = \\ = \frac{ \cos {}^{2} ( \beta ) ( \frac{ \sin( \beta ) }{ \cos( \beta ) } + \frac{2 \sin {}^{2} ( \beta ) }{ \cos {}^{2} ( \beta ) } + 2)}{ \cos {}^{2} ( \beta )(6 + \frac{ \sin {}^{2} ( \beta ) }{ \cos {}^{2} ( \beta ) } } = \\ = \frac{tg \beta + 2 {tg}^{2} \beta + 2 }{6 + {tg}^{2} \beta } = \\ \\ = \frac{2 + 2 \times 4 + 2}{6 + 4} = \frac{12}{10} = 1.2$

2.

$\frac{ \sin( \beta ) \cos( \beta ) - 3}{6 \cos {}^{2} ( \beta ) - \sin {}^{2} ( \beta ) } = \\ = \frac{ \sin( \beta ) \cos( \beta ) - 3 \sin {}^{2} ( \beta ) - 3 \cos {}^{2} ( \beta ) }{ 6\cos {}^{2} ( \beta ) - \sin {}^{2} ( \beta ) } = \\ = \frac{ \cos {}^{2} ( \beta )( \frac{ \sin( \beta ) }{ \cos( \beta ) } - 3 \frac{ \sin {}^{2} ( \beta ) }{ \cos {}^{2} ( \beta ) } - 3) }{ \cos {}^{2} ( \beta ) (6 - \frac{ \sin {}^{2} ( \beta ) }{ \cos {}^{2} ( \beta ) } )} = \\ = \frac{tg \beta - 3 {tg}^{2} \beta - 3 }{6 - {tg}^{2} \beta } = \\ \\ = \frac{ - 2 - 12 - 3}{6 - 4} = \frac{ - 17}{2} = - 6.5$

3.

$\frac{2 \sin( \beta ) \cos( \beta ) + 3} {4 \cos {}^{2} ( \beta ) + \sin {}^{2} ( \beta ) } = \\ = \frac{2 \sin( \beta ) \cos( \beta ) + 3 \sin {}^{2} ( \beta ) + 3\cos {}^{2} ( \beta ) }{4 \cos {}^{2} ( \beta ) + \sin {}^{2} ( \beta ) } = \\ = \frac{2tg \beta + 3 {tg}^{2} \beta + 3 }{4 + {tg}^{2} \beta } = \\ \\ = \frac{ - 8 + 3 \times 16 + 3}{4 + 16} = \frac{ - 5 + 48}{20} = \\ = \frac{43}{20}$

4

$\frac{ \cos {}^{2} ( \beta ) + 2 }{ \cos {}^{2} ( \beta ) + 3 \sin( \beta ) \cos( \beta ) } = \\ = \frac{ \cos {}^{2} ( \beta ) + 2\sin {}^{2} ( \beta ) + 2\cos {}^{2} ( \beta ) }{ \cos {}^{2} ( \beta ) + 3 \sin( \beta ) \cos( \beta ) } = \\ = \frac{2 \sin {}^{2} ( \beta ) + 3\cos {}^{2} ( \beta ) }{ \cos {}^{2} ( \beta ) + 3 \sin( \beta ) \cos( \beta ) } = \\ = \frac{2 {tg}^{2} \beta + 3}{1 + 3tg \beta } = \\ \\ = \frac{2 \times 9 + 3}{1 + 9} = \frac{21}{10} = 2.1$