Consider the angle sum formulas: #sin(A+B)=sinA cosB+cosA sinB# #cos(A+B)=cosA cosB−sinA sinB# show that #tan(90^0+θ)=− 1/tanθ#. and use this identity to prove that lines #y = mx and y = m_1x# are perpendicular iff #mm_1=−1#?

1 Answer
Jan 8, 2017

Given

#sin(A+B)=sinAcosB+cosAsinB....[1]#

#cos(A+B)=cosAcosB-sinAsinB....[2]#

So #tan(A+B)=sin(A+B)/cos(A+B)#

#=((sinAcosB)/(cosAcosB)+(cosAsinB)/(cosAcosB))/((cosAcosB)/(cosAcosB)-(sinAsinB)/(cosAcosB))#

#=>tan(A+B)=(tanA+tanB)/(1-tanAtanB).....[3]#

#=>tan(A+B)=(tanA/tanA+tanB/tanA)/(1/tanA-tanB)#

#=>tan(A+B)=(1+tanB*cotA)/(cotA-tanB).....[4]#

This is an identity so it is valid for real values of #A and B#.So putting #A=90^@ and B= theta # in [4] we get

#=>tan(90^@+theta)=(1+tantheta*cot90)/(cot90-tantheta)#

#=>tan(90^@+theta)=(1+tantheta*0)/(0-tantheta)=-1/tantheta#

#=>tan(90^@+theta)=-1/tantheta....[5]#

Now if #y=mx# straight line makes angle #theta# with positive direction of x-axis then #m=theta#.Again if the straight line having equation #y=m_1x# makes an angle #90^@+theta# with the positive direction of x-axis,then it will be perpendicular to the first straight line and its slope m_1=tan(90^@+theta#

So by relation [5] we get

#m_1=-1/m#

#mm_1=-1#, for mutually perpendicular straight lines