Question

If sin^2A+sin^2B+sin^2C=3 then find the value of cos^2A+cos^2B+cos^2C​

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