If tan theta = 12/5 and cos theta < 0, what is sin theta ?
a.12
b.5/13
c.-5/13
d.12/13
e.-12/13

1 + tan ² Ф=sec²Ф
1+(12/5)²=sec²Ф
169/25=sec² Ф
sec Ф=⁺₋√(169/25)=⁺₋13/5

sec Ф=1/cos Ф ⇒cosФ=1/sec Ф
 cos Ф>0   ⇔  sec Ф>0  ⇔  sec Ф=+ 13/5

cos Ф=1/secФ
cos Ф=1 / 13/5=5/13

we can calculate the sin Ф, with this method.
sin²Ф + cos²Ф=1  ⇒ sin Ф=⁺₋√(1-cos² Ф)
sin Ф=⁺₋√[1-(5/13)²]=⁺₋12/13
like cos Ф>0 and tan Ф>0   ⇒ sin Ф>0 ⇒sin Ф=12/13

d.12/13

other method
tan Ф=sin Ф / cos Ф
12/5=sin Ф / 5/13
sin Ф=(12/5)*(5/13)=12/13

d.12/13

The expressions would be 0.25a

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ),   by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i   0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5