Question

If sec theta=13/12 find cos theta sin theta​

Answer 1

Answer:

$\sec \alpha = 13 \div 12 \\ \sec \alpha = h \div b \\ h = 13 \: b = 12 \\ p = \sqrt{h {}^{2} } - b {}^{2} \\ p = \sqrt{13 {}^{2} } - 12 { }^{2} \\ p = \sqrt{169 - 144 } \\ p = \sqrt{25} \\ p = 5 \\ \\ \cos( \alpha ) = b \div h \\ = 12 \div 13 \\ \sin( \alpha ) = p \div h \\ = 5 \div 13$

$\cos( \alpha ) = b \div h \\ = 12 \div 13 \\ \sin( \alpha) = p \div h \\ = 5 \div 13$

Answer 2

$\sec \: theta = \frac{13}{12} \\ cos \: theta = \frac{1}{sec \: theta} \\ \: \: \: \: \: = \frac{12}{13} \\ sin \: theta = \sqrt{1 - {cos}^{2} theta} \\ \: \: \: \: \: \: \: \: \: \: \: = \sqrt{1 - (\frac{12}{13}}) ^{2} \\ = \sqrt{1 - \frac{144}{169} } \\ = \sqrt{ \frac{25}{169} } \\ = \frac{5}{13}$

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