How do you find the other five trigonometric values given csc b = -6/5 and cot b > 0?

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Answer 1 · 2016-07-03T08:02:57

#sinb=-5/6#, #cosb=-sqrt11/6#, #tanb=5/sqrt11#,

#cotb=sqrt11/5# and #secb=-6/sqrt11#.

As #cscb=-6/5# #=># #csc^2b=36/25# and using #csc^2b=1+cot^2b#, we have

#1+cot^2b=36/25# and #cotb=sqrt(36/25-1)=sqrt11/5# (as #cotb>0#)

As #cotb=sqrt11/5#, #tanb=1/cotb=5/sqrt11#

As #cscb=-6/5#, #sinb=1/cscb=-5/6#

and #cosb=cosb/sinbxxsinb=cotbxxsinb#

= #sqrt11/5xx(-5/6)=-sqrt11/6#

and #secb=1/cosb=-6/sqrt11#.

Hence, we have #sinb=-5/6#, #cosb=-sqrt11/6#, #tanb=5/sqrt11#,

#cotb=sqrt11/5# and #secb=-6/sqrt11#.