Question

A+B+C=90 then prove that sinC.cosC+sinB.cosB+sinA.cosA=2cosA.cosB.cosC

Answer 1

Answer:

this is the answer for the question

Answer 2

Proof:

$Given\,\,that\\ A+B+C =\pi/2\\\implies A+B=\pi/2-C\,\,.........(1)$

LHS

$= sinAcosA +sinBcosB +sinCcosC\\=\frac{1}{2} \bigg[2sin\left(\frac{2A+2B}{2} \right)cos\left(\frac{2A-2B}{2} \right) +2sinCcosC\bigg]\\=sin(A+B)cos(A-B)+sinC cosC\\=sin(\pi/2-C)(cosAcosB + sinAsinB)+sin(\pi/2-(A+B))cosC\\=cosC[cos(A-B)+cos(A+B)]\\=2cosAcosBcosC$

=RHS

Hence Proved.

Properties and results used:

1.sin2A = 2sinAcosA

2.sinC +sinD=2Sin[(C+D)/2]Cos[(C-D)/2]

3.Cos(A+B)+Cos(A-B) = 2CosACosB

4.sin(90-A)=cosA