Question

LHS=cos²A/2+cos²B/2-cos²C/2
RHS=2cosA/2cosB/2sinC/2

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Answer 1

Answer:

A+B+C=π⇒B+C=π−A

Given, cos

2

A+cos

2

B+cos

2

C=1

⇒cos

2

A+cos

2

B=1−cos

2

C=sin

2

C

⇒−cos

2

A=cos

2

B−sin

2

C=cos(B+C).cos(B−C)=−cosA.cos(A−C)

⇒cosA(cosA−cos(B−C))=0

⇒cosA(cos(π−(B+C)−cos(B−C))=0

⇒cosA(−cos(B+C)−cos(B−C))=0

⇒cosA(2cosBcosC)=0

∴ either cosA=90

o

or cosB=90

o

or cosC=90

o

So, its a triangle with one angle =90

o

Hence, right-angled triangle