Consider the angle sum formulas: $sin (A + B) = sin A cos B + cos A sin B$ $sin(A+B)=sinA cosB+cosA sinB$ $cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B)=cosA cosB−sinA sinB$ 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 $mm_1=−1$ ?

Question

Consider the angle sum formulas: $sin (A + B) = sin A cos B + cos A sin B$ $sin(A+B)=sinA cosB+cosA sinB$ $cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B)=cosA cosB−sinA sinB$ 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

Impact of this question

5190 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-08T08:26:42

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

Science

Math

Humanities

... and beyond

1 Answer

Related questions

Impact of this question