tentukan cos teta dan tan teta dalam x jika diketahui a. sin teta=2x b. csc teta=x+1 c.sec teta=3x+1

[tex]a. \sin( \alpha ) =2x \\ { \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1 \\ \cos ^{2} ( \alpha ) = 1 - 4 {x}^{2} \\ \cos ( \alpha ) = \sqrt{1 - 4 {x}^{2} } \\ \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{2x}{ \sqrt{1 - {4x}^{2} } } = \frac{2x \sqrt{1 - 4 {x}^{2} } }{1 - 4 {x}^{2} } \\ \\
b. \csc( \alpha ) =x+1 \\ \frac{1}{ \sin( \alpha ) } = x + 1 \\ \sin( \alpha ) = \frac{1}{x + 1} \\ { \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1 \\ { \cos( \alpha ) }^{2} = 1 - \frac{1}{ {x}^{2} + 2x + 1} = \frac{ {x}^{2} + 2x }{ {x}^{2} + 2x + 1 } \\ \cos( \alpha ) = \sqrt{ \frac{ {x }^{2} + 2x }{ {x}^{2} + 2x + 1 } } = \frac{ \sqrt{ {x}^{2} + 2x } }{x + 1} \\ \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{ \frac{1}{x + 1} }{ \frac{ \sqrt{ {x}^{2} + 2x } }{x + 1} } = \frac{1}{ \sqrt{ {x}^{2} + 2x } } = \frac{ \sqrt{ {x}^{2} + 2x } }{ {x}^{2} + 2x} \\ \\
c. \sec( \alpha ) =3x+1 \\ \frac{1 }{ \cos( \alpha ) } = 3x + 1 \\ \cos( \alpha ) = \frac{1}{3x + 1} \\ { \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1 \\ { \sin( \alpha ) }^{2} = 1 - \frac{1}{9 {x}^{2} + 6x + 1} = \frac{9 {x}^{2} + 6x}{9 {x}^{2} + 6x + 1} \\ \sin( \alpha ) = \sqrt{\frac{9 {x}^{2} + 6x}{9 {x}^{2} + 6x + 1}} = \frac{ \sqrt{9 {x}^{2} + 6x } }{3x + 1} \\ \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{\frac{ \sqrt{9 {x}^{2} + 6x } }{3x + 1}}{ \frac{1}{3x + 1} } = \sqrt{9 {x}^{2} + 6 } [/tex]