Question

in a right angle triangle it is given that angle A is an acute angle and that tan A equals to 5 by 12 find the values of cos to cosec a - cot a​

Answer 1

Answer:

$\tan(a) = \frac{5}{12} \\ \cot(a) = \frac{12}{5} \\ 1.we \: know \: that \\ { \tan(a) }^{2} + 1 = { \sec(a) }^{2} \\ { \sec(a) }^{2} = { (\frac{5}{12}) }^{2} + 1 \\ { \sec(a) }^{2} = \frac{25}{144} + 1 \\ { \sec(a) }^{2} = \frac{25 + 144}{144} \\ { \sec(a) }^{2} = \frac{169}{144} \\ \sec(a) = \sqrt{ \frac{169}{144} } \\ \sec(a ) = \frac{13}{12} \\ \cos(a) = \frac{12}{13} \\ 2.we \: know \: that \\ \ { \sin(a) }^{2} + { \cos(a) }^{2} = 1 \\ { \sin(a) }^{2} = 1 - { \cos(a) }^{2} \\ { \sin(a) }^{2} = 1 - { (\frac{12}{13} )}^{2} \\ { \sin(a) }^{2} = 1 - \frac{144}{169} \\ { \sin(a) }^{2} = \frac{169 - 144}{169} \\ { \sin(a) }^{2} = \frac{25}{169} \\ \sin(a) = \sqrt{ \frac{25}{169} } \\ \sin(a) = \frac{5}{13} \\ \cosec(a) = \frac{13}{5} \\ \cosec(a) - \cot(a) = \frac{13}{5} - \frac{12}{5} \\ = \frac{1}{5}$

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