Tan-1-a-bca-b-c-tan-1-b-aca-b-c-tan-1-c-aca-b-c-

Θ =tan^ 1ar+tan^ 1br+tan^ 1cr Where r =√({(a+b+c)/abc}) =π +tan^ 1ar+br/1 arbr+tan^ 1cr ∴ 1 abr^2 =1 (a+b+c)/c[∵ a,b,c gt; 0] = (a+b)/c lt;0 =π +Tan^ 1 ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

tan-1√a/bca+b...

Question

$t a n^{- 1} \sqrt{\frac{a}{b c} (a + b + c)} + t a n^{- 1} \sqrt{\frac{b}{a c} (a + b + c)} + t a n^{- 1} \sqrt{\frac{c}{a c} (a + b + c)} =$

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π

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- π

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None

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Solution

The correct option is B $π$
$θ = t a n^{- 1} a r + t a n^{- 1} b r + t a n^{- 1} c r$
Where $r = \sqrt{{\frac{(a + b + c)}{a b c}}}$
$= π + t a n^{- 1} \frac{a r + b r}{1 - a r b r} + t a n^{- 1} c r ∴ 1 - a b r^{2} = 1 - \frac{(a + b + c)}{c} [∵ a, b, c > 0] = - \frac{(a + b)}{c} < 0 = π + T a n^{- 1} (- c r) + t a n^{- 1} c r = π - t a n^{- 1} c r + t a n^{- 1} c r = π$

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tan−1√abc(a+b+c)+tan−1√bac(a+b+c)+tan−1√cac(a+b+c)=

The correct option is B πθ=tan−1ar+tan−1br+tan−1cr Where r=√{(a+b+c)abc} =π+tan−1ar+br1−arbr+tan−1cr∴1−abr2=1−(a+b+c)c[∵a,b,c>0]=−(a+b)c<0=π+Tan−1(−cr)+tan−1cr=π−tan−1cr+tan−1cr=π

$t a n^{- 1} \sqrt{\frac{a}{b c} (a + b + c)} + t a n^{- 1} \sqrt{\frac{b}{a c} (a + b + c)} + t a n^{- 1} \sqrt{\frac{c}{a c} (a + b + c)} =$