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If sin^2A+sin^2B+sin^2C=3 then find the value of cos^2A+cos^2B+cos^2C |
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Answer» A+B+C=π⇒B+C=π−AGiven, 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 |
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