In a triangle ABC, cos2A+cos2B+cos2C=1. Then the triangle is
a)right angled
b) obtuse angled
c)acute angled
d)equilateral
1cos2A+cos2B+cos2C
=1-sin2A+cos2B+cos2C
=1+(cos2B-sin2A)+cos2C
=1+cos(B+A)cos(B-A)+cos2C
=1-cosC cos(B-A)+cos2C
=1-cosC[cos(B-A)-cosC]
=1-cosC[cos(B-A)+cos(B+A)]
=1-cosC[2cosAcosB]
=1-2cosAcosBcosC
BY THE EQUATION,THEN
1-2cosAcosBcosC=1
THEN
2cosAcosBcosc=0
THEN EITHER cosA=0 OR cosB=0 OR cosC=0
SO EITHER angle (A or B or C)=900
so option (a) triangle is right angled is correct
49indeed a genius solution...[1][1][1]
but by observation it comes a right angled triangle.......
coz suppose A=90°
then B=90-C
thus cos2A+cos2B+cos2C=cos2(90)+cos2(90-C)+cos2C
=(0)2+sin2C+cos2C=1
which satisfies the given results........thus an equilateral triangle.....[4][4][4][4]
