Question

tan theta = -2 fintand the value of cos theta and cosec theta when theta lies in the second quadrant​

Answer 1

Answer:

$\tan( \alpha ) = - 2 = \frac{p}{b}$

$h = \sqrt{ {p}^{2} + {b}^{2} } = \sqrt{ {2}^{2} + {1}^{2} } = \sqrt{5}$

in 2nd quadrant, cos -ve & cosec +ve.

$\cos( \alpha ) = \frac{b}{h} = \frac{ - 1}{ \sqrt{5} }$

$\csc( \alpha ) = \frac{h}{p} = \frac{ \sqrt{5} }{2}$