Question

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`?

Answer 1

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