Solve into simplest form-tan -1a-b-1-a b-tan -1b-c-1-b c

Tan^ 1(a b/1+ab )+tan^ 1(b c/1+bc ) Let a = tan x b = tan y c = tan z ∴ tan^ 1(tan y tan y/1+tan x. tan y )+tan^ 1(tan y tan z/1+tan y. tan z ) =tan^ 1(tan ...

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Standard XI

Mathematics

Equations with Solutions at Boundary Values

Solve into si...

Question

Solve into simplest form:
$t a n^{- 1} (\frac{a - b}{1 + a b}) + t a n^{- 1} (\frac{b - c}{1 + b c})$

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Solution

$t a n^{- 1} (\frac{a - b}{1 + a b}) + t a n^{- 1} (\frac{b - c}{1 + b c})$
Let a = tan x
b = tan y
c = tan z
$∴ t a n^{- 1} (\frac{t a n y - t a n y}{1 + t a n x . t a n y}) + t a n^{- 1} (\frac{t a n y - t a n z}{1 + t a n y . t a n z})$
$= t a n^{- 1} (t a n (x - y)) + t a n^{- 1} (t a n (y - z)) = x - y + y - z = x - z = t a n^{- 1} a - t a n^{- 1} c$

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Similar questions

Evaluate tan^-1(a - b)/(1+ab)+ tan^-1(b - c)/(1+bc)+ tan^-1(c - a)/(1+cb) ; ab,bc,ac>-1

Q. Prove that:

{tan}^{- 1} (\frac{a - b}{1 + a b}) + {tan}^{- 1} (\frac{b - c}{1 + b c}) + {tan}^{- 1} (\frac{c - a}{1 + c a}) = 0

Q. Assertion :If

a > b > 0

a c < - 1 < b c

, then

{tan}^{- 1} (\frac{a - b}{1 + a b}) + {tan}^{- 1} (\frac{b - c}{1 + b c}) + {tan}^{- 1} (\frac{c - a}{1 + a c}) = π

Reason:

{tan}^{- 1} x - {tan}^{- 1} y = {tan}^{- 1} (\frac{x - y}{1 + x y}) \forall x, y

$t a n^{- 1} a + t a n^{- 1} b$ , where a>0,b>0,ab>1, is equal to

{tan}^{- 1} a + {tan}^{- 1} b,

where

a > 0, b > 0, a b > 1,

is equal to

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Solve into simplest form: tan−1(a−b1+ab)+tan−1(b−c1+bc)

tan−1(a−b1+ab)+tan−1(b−c1+bc) Let a = tan x b = tan y c = tan z ∴tan−1(tany−tany1+tanx.tany)+tan−1(tany−tanz1+tany.tanz) =tan−1(tan(x−y))+tan−1(tan(y−z))=x−y+y−z=x−z=tan−1a−tan−1c

Solve into simplest form:
$t a n^{- 1} (\frac{a - b}{1 + a b}) + t a n^{- 1} (\frac{b - c}{1 + b c})$