If #sectheta+tantheta=3/2#, what is the value of #sintheta#?

Question

Asher Friedman · Accepted Answer

#sin theta=5/13# #tan theta +sec theta =3/2# No idea-well do something!! #sin theta/cos theta+1/cos theta=3/2# And then the next stage becomes clear #(sin theta+1)/cos theta=3/2# #2(sin theta +1)=3cos theta# So what now.? Well we only want to know about #sin theta# And we do know #sin^2 theta +cos^2 theta=1# So square both sides #4(sin theta+1)^2=9cos^2 theta# #4(sin^2 theta +2sin theta +1)=9(1-sin^2 theta)# #4sin^2 theta +8sin theta+4=9-9 sin^2 theta# #13sin^2theta+8sin theta-5=0# Factorise #(13 sin theta-5)(sintheta +1)=0# #sintheta=5/13# or #sintheta=-1# Only the first will do because if #sintheta =-1# then #costheta=0# and clearly we cannot divide by zero.

Madelyn Adams · Answer

Given

#sectheta+tantheta=1.5=15/10=3/2#

#=>sectheta+tantheta=3/2……………(1)#

Again we know

#sec^2theta-tan^2theta=1……………(2)#

Dividing (2) by (1) we get

#sectheta-tantheta=2/3……………(3)#

Adding (1) and (3) we get

#2sectheta=3/2+2/3=13/6#

#=>sectheta=13/12#

Subtracting (3) from (1) we get

#2tantheta=3/2-2/3=5/6#

#=>tantheta=5/12#

#=>sinthetaxxsectheta=5/12#

#=>sinthetaxx13/12=5/12#

#=>sintheta=5/12xx12/13=5/13#

Alternative

#sectheta+tantheta=3/2#

#=>1/costheta+sintheta/costheta=3/2#

#=>(1+sintheta)/costheta=3/2#

#=>(1+sintheta)/sqrt(1-sin^2theta)=3/2#

#=>(sqrt(1+sintheta)sqrt(1+sintheta))/(sqrt(1-sintheta)sqrt(1+sintheta))=3/2#

for #sintheta!=-1#

#=>(sqrt(1+sintheta))/(sqrt(1-sintheta))=3/2#

#=>(1+sintheta)/(1-sintheta)=9/4#

#=>4+4sintheta=9-9sintheta#

#=>13sintheta=5#

#=>sintheta=5/13#