Question

cos^(-1)a+cos^(-1)b+cos^(-1)c=π,.. prove that a^2+b^2+c^2+2abc=1.......please answer with solutions.

Answer 1

hope this will helps you

Answer 2

it is given that, cos-¹a + cos-¹b + cos-¹c = π

we have to prove that a² + b² + c² + 2abc = 1

cos-¹a + cos-¹b + cos-¹c = π

cos-¹a + cos-¹b = π - cos-¹c

⇒cos-¹(ab - √(1 - a²).√(1 - b²)) = π - cos-¹c

⇒ab - √(1 - b² - a² + a²b²) = cos(π - cos-¹c)

= ab - √(1 - b² - a² + a²b²) = -cos(cos-¹c)

⇒ab - √(1 - b² - a² + a²b²) = -c

⇒ab + c = √(1 - b² - a² + a²b²)

⇒a²b² + c² + 2abc = 1 - b² - a² + a²b²

⇒a² + b² + c² + 2abc = 1 [hence proved]

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