Question

cosA/sinB-cosA.SinB=sin 2 A if A+B=90°​

Answer 1

Answer:

cos(A+B)=cosAcosB−sin AsinB....[2] ... =sinAcosBcosAcosB+cosAsinB cosAcosBcosAcosBcosAcosB− ... real values of AandB .So putting A=90∘andB=θ in [4] we get.

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#?

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

Consider the angle sum formulas:

. and use this identity to prove that lines y = mx and y = m_1x

Substituting the values of cos450, cos300, sin450, and sin300, we get ... cosA cosB – sinA sinB; sin(A – B) = cos[900 – (A – B)] ... We know that, sin(A + B) + sin(A – B) = 2 sinA cosB

Therefore both has to be equal to 1. SinA=SinB =1,which implies A=B=90 degrees. Cos(90)=0. Therefore, Cos A+Cos B= ...