What is #tan(arcsin(12/13)) #?

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Answer 1 · 2015-07-21T21:32:44

#tan(arcsin(12/13))=12/5#

Let#" "theta=arcsin(12/13)#

This means that we are now looking for #color(red)tantheta!#

#=>sin(theta)=12/13#

Use the identity,

#cos^2theta+sin^2theta=1#

#=>(cos^2theta+sin^2theta)/cos^2theta=1/cos^2theta#

#=>1+sin^2theta/cos^2theta=1/cos^2theta#

#=>1+tan^2theta=1/cos^2theta#

#=> tantheta=sqrt(1/cos^2(theta)-1)#

Recall : #cos^2theta= 1-sin^2theta#

#=>tantheta=sqrt(1/(1-sin^2theta)-1)#

#=>tantheta=sqrt(1/(1-(12/13)^2)-1)#

#=>tantheta=sqrt(169/(169-144)-1#

#=>tantheta=sqrt(169/25-1)#

#=>tantheta=sqrt(144/5)= 12/5#

REMEMBER what we called #theta# was actually #arcsin(12/13)#

#=>tan(arcsin(12/13))= color(blue)(12/5)#