If a- b- c are positive - then - tan-1-aa-b-c-bc-

The correct option is A π Let A=√(a(a+b+c)/bc) ABC =a+b+c/abc A+B+C=√(a+b+c/abc)(a+b+c) AB=a+b+c/c BC=a+b+c/a → tan ^ 1A+ta ...

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Standard XII

Mathematics

Algebra of Derivatives

If a, b, c ...

Question

If $a, b, c$ are positive , then $\sum t a n^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} =$

π

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\frac{3 π}{2}

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\frac{3 π}{4}

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3

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Solution

The correct option is A $π$
Let $A = \sqrt{\frac{a (a + b + c)}{b c}}$
$A B C = \frac{a + b + c}{a b c}$
$A + B + C = \sqrt{\frac{a + b + c}{a b c}} (a + b + c)$
$A B = \frac{a + b + c}{c}$
$B C = \frac{a + b + c}{a}$
$\to t a n^{- 1} A + t a n^{- 1} B + t a n^{- 1} C$
$= t a n^{- 1} (\frac{A + B}{1 - A B}) + t a n^{- 1} C$
$= t a n^{- 1} (\frac{A + B}{1 - A B}) + t a n^{- 1} C$
$t a n^{- 1} \frac{(\frac{A + B}{1 - A B} + C)}{1 - (\frac{A C + B C}{1 - A B})}$
$= t a n^{- 1} (\frac{A + B + C - A B C}{1 - A B - B C - A C})$
$= t a n^{- 1} \frac{(\frac{a + b + c}{a b c} - (a + b + c) \sqrt{\frac{a + b + c}{a b c}})}{1 - (a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})}$
$= t a n^{- 1} (0)$
$= π$

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If a, b, c are positive , then ∑tan−1√a(a+b+c)bc=

If $a, b, c$ are positive , then $\sum t a n^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} =$