Prove that-tan -12 aba 2- b 2-tan -12 xyx 2- y 2-tan -12 - 2- 2where -a x-b y and -a y-b x

We know tan 1x+tan 1y=tan 1x+y1 xy, #160; #160;xy #62;1 #8756; #160;tan 12aba2 b2+tan 12xyx2 y2=tan 12aba2 b2+2xyx2 y21 2aba2 b22xyx2 y2=tan 12abx2 ...

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Standard XII

Mathematics

Derivative

Prove that:ta...

Question

Prove that:
$\tan^{- 1} \frac{2 a b}{a^{2} - b^{2}} + \tan^{- 1} \frac{2 x y}{x^{2} - y^{2}} = \tan^{- 1} \frac{2 αβ}{α^{2} - β^{2}},$
where α = ax − by and β = ay + bx.

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Solution

We know
$\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y}), x y > 1$

$∴ \tan^{- 1} \frac{2 a b}{a^{2} - b^{2}} + \tan^{- 1} \frac{2 x y}{x^{2} - y^{2}} = \tan^{- 1} (\frac{\frac{2 a b}{a^{2} - b^{2}} + \frac{2 x y}{x^{2} - y^{2}}}{1 - \frac{2 a b}{a^{2} - b^{2}} \frac{2 x y}{x^{2} - y^{2}}}) = \tan^{- 1} (\frac{\frac{2 (a b x^{2} - a b y^{2} + x y a^{2} - x y b^{2})}{(a^{2} - b^{2}) (x^{2} - y^{2})}}{\frac{(a^{2} x^{2} - a^{2} y^{2} - x^{2} b^{2} + y^{2} b^{2} - 4 a b x y)}{(a^{2} - b^{2}) (x^{2} - y^{2})}}) = \tan^{- 1} (\frac{2 (a b x^{2} - a b y^{2} + x y a^{2} - x y b^{2})}{(a^{2} x^{2} - a^{2} y^{2} - x^{2} b^{2} + y^{2} b^{2} - 2 a b x y - 2 a b x y)}) = \tan^{- 1} (\frac{2 (a x - b y) (a y + b x)}{{(a x - b y)}^{2} - {(a y + b x)}^{2}}) = \tan^{- 1} (\frac{2 α β}{α^{2} - β^{2}}) [∵ α = a x - b y and β = a y + b x]$

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Prove that: tan-12aba2-b2+tan-12xyx2-y2=tan-12αβα2-β2, where α = ax − by and β = ay + bx.

Prove that:
$\tan^{- 1} \frac{2 a b}{a^{2} - b^{2}} + \tan^{- 1} \frac{2 x y}{x^{2} - y^{2}} = \tan^{- 1} \frac{2 αβ}{α^{2} - β^{2}},$
where α = ax − by and β = ay + bx.