Question

13
1) If sec thita =13/12
find tan thita​

Answer 1

$\pink{given : } \\ \\ \sec( \alpha ) = \frac{13}{12} = \frac{hypotenuse}{adjustent} \\ \\ \blue{to \: \: \: find \: \: : } \tan( \alpha ) \\ \\ \red{solution : } \\ \\ opposite = \sqrt{ {hypt}^{2} - {adjut}^{2} } \\ \\ oppo = \sqrt{ {13}^{2} - {12}^{2} } \\ \\ oppo \: = \sqrt{169 - 144} \\ \\ oppo = \sqrt{25} \\ \\ oppo = 5 \\ \\ \pink{hence} \\ \\ \tan( \alpha ) = \frac{oppo}{adjust} \\ \\ \tan( \alpha ) = \frac{5}{12}$

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Answer 2

Answer:

tan thita = 5/12

Step-by-step explanation:

sec thita = K/A = 13/12

K= 13. A= 12

We know that

K^2 = A^2 + L^2

L^2 = K^2 - A^2

L^2 = (13)^2 - (12)^2

L^2 = 169 - 144

L^2 = 25

L = 5

then , we know that

tan thita = L/A

tan thita = 5/12