Question

Спростіть виразооолллдддддд

Answer 1

1)

$1 - \cos {}^{2} ( \alpha ) = \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) - \cos {}^{2} ( \alpha ) = \sin {}^{2} ( \alpha )$

2)

$\sin {}^{2} ( \beta ) - 1 = \sin {}^{2} ( \beta ) - ( \sin {}^{2} ( \beta ) + \cos {}^{2} ( \beta ) ) = \\ = \sin {}^{2} ( \beta ) - \sin {}^{2} ( \beta ) - \cos {}^{2} ( \beta ) = - \cos {}^{2} ( \beta )$

3)

$\sin {}^{2} (q) + \cos {}^{2} (q) + 1 = 1 + 1 = 2$

4)

$\sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1 \: \: | \div \cos {}^{2} ( \alpha ) \\ \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } + \frac{ \cos {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } = \frac{1}{ \cos {}^{2} ( \alpha ) } \\ \frac{1}{ \cos {}^{2} ( \alpha ) } = \tan {}^{2} ( \alpha ) + 1 \\ \\ \frac{1}{ \cos {}^{2} ( \alpha ) } - 1 = \tan {}^{2} ( \alpha ) + 1 - 1 = \tan {}^{2} ( \alpha )$

Answer 2

Ответ:

Применяем основное тригонометрическое тождество  

        $\bf sin^2a+cos^2a=1$  .

$\bf 1)\ \ 1-cos^2a=(\underbrace{\bf sin^2a+cos^2a}_{1})-cos^2a=sin^2a+cos^2a-cos^2a=sin^2a\\\\\\2)\ \ sin^2\beta -1=sin^2\beta -(\underbrace{\bf sin^2\beta +cos^2\beta }_{1})=sin^2\beta -sin^2\beta -cos^2\beta =-cos^2\beta \\\\\\3)\ \ \underbrace{\bf sin^2\varphi+cos^2\varphi }_{1}+1=1+1=2\\\\\\4)\ \ \dfrac{1}{cos^2\varphi }-1=\dfrac{\overbrace{\bf sin^2\varphi +cos^2\varphi }^{1}}{cos^2\varphi }-1=\dfrac{sin^2\varphi }{cos^2\varphi }+\dfrac{cos^2\varphi }{cos^2\varphi }-1=tg^2\varphi +1-1=tg^2\varphi$