Question

If cot θ = -3/2 , find cos θ.

Answer 1

Answer:

cot $=b/p

cos $=b/h

=-3/(13)^1/2

Answer 2

Given,

$\cot( \alpha ) = \frac{ - 3}{2}$

We know,

$\tan( \alpha ) = \frac{1}{ \cot( \alpha ) }$

Thus,

$\tan( \alpha ) = \frac{ - 2}{3}$

Also,

${ \sec( \alpha ) ) }^{2} - { \tan( \alpha ) }^{2} = 1 \\ = > { \sec( \alpha ) }^{2} = 1 + { \frac{ - 2}{ {3}^{2} } }^{2} \\ = \frac{13}{9} = \\ \sec( \alpha ) = \frac{ \sqrt{13} }{3}$

Hence ,

$\cos( \alpha ) = \frac{3}{ \sqrt{13} } )$

Regards

Kshitij