Question

from the adjacent figure
i)prove that sin²alpha+cos²alpha=sin²theta+cos²theta
ii) prove that sec²alpha-tan²alpha=sec²theta-tan²theta​

Answer 1

Step-by-step explanation:

$pr = \sqrt{ {13}^{2} - {5}^{2} } \\ \sin(theta) = 12 \: cm \\ \sin(theta) = \frac{12}{13} \\ \\ \cos(theta) = \frac{5}{13} \\ \\ \sin( \alpha ) = \frac{5}{13} \\ \cos( \alpha ) = \frac{12}{13} \\ \\ therefore \: \: \: \: { (\sin(theta)) }^{2} + {( \cos(theta)) }^{2} = \frac{ {12}^{2} }{ {13}^{2} } + \frac{ {5}^{2} }{ {13}^{2} } = \frac{ {12}^{2} + {5}^{2} }{ {13}^{2} } \\ = \frac{144 + 25}{169} = \frac{169}{169} = 1 \\ \\ { (\sin( \alpha )) }^{2} + { (\cos( \alpha )) }^{2} = \frac{ {5}^{2} }{ {13}^{2} } + \frac{ {12}^{2} }{13} = \frac{25 + 144}{169} = \frac{169}{169} = 1 \\ \\ therefore \: \: \: \: {( \sin(theta)) }^{2} + {( \cos(theta)) }^{2} = {( \sin( \alpha )) }^{2} + {( \cos( \alpha )) }^{2} \\ \\ \\ \\ \\ \tan(theta) = \frac{12}{5} \\ \\ \sec(theta) = \frac{13}{5} \\ \\ \tan( \alpha ) = \frac{5}{12} \\ \sec( \alpha ) = \frac{13}{12} \\ \\ so \: \: \: \: \: \: \: \: { \sec(theta) }^{2} - { \tan(theta) }^{2} = \frac{ {13}^{2} }{ {5}^{2} } - \frac{ {12}^{2} }{ {5}^{2} } = \frac{ {169 - 144} }{25} \\ = \frac{25}{25} = 1 \\ \\ \\ \\ {( \sec( \alpha )) }^{2} - {( \tan( \alpha )) }^{2} = \frac{ {13}^{2} }{ {12}^{2} } - \frac{ {5}^{2} }{ {12}^{2} } = \frac{169 - 25}{144} = \frac{144}{144} = 1 \\ \\ \\ \\ therefore \: \: \: \: { (\sec(theta)) }^{2} - {( \tan(theta) )}^{2} = {( \sec( \alpha ) )}^{2} - {( \tan( \alpha )) }^{2}$