Question

180°<Φ<270° and sinΦ=-5/13 then 5cot²+12tanΦ+13cosecΦ​

Answer 1

Answer:

Perpendicular = 5

Hypotenuse = 13

Then, base = - 12

$\sqrt{ {13}^{2} + {5}^{2} } = 12$

Cot = B/P

Tan = P/B

Cosec = H/P

So,

$5 \times ( { \frac{12}{5} )}^{2} - 12 \times \frac{5}{12} + 13 \times \frac{13}{5} \\ = \frac{144}{5} - 5 + \frac{169}{5} \\ = \frac{313}{5} - 5 = \frac{313 - 25}{5} = \frac{288}{5}$