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^{ -1 }{ [ { { 2y+1+2x+1 }{ { 3 } } }{ 1-( { 4xy+2x2y+1

Question

$tan^{- 1} [\frac{\frac{2 y + 1 + 2 x + 1}{3}}{1 - ( \frac{4 x y + 2 x 2 y + 1}{3} )}] = k$

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tan^{- 1} [\frac{\frac{2 y + 1 + 2 x + 1}{3}}{1 - ( \frac{4 x y + 2 x 2 y + 1}{3} )}] = k

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Text solutionVerified

Consider

\frac{\frac{2 y + 1 + 2 x + 1}{3}}{1 - ( \frac{4 x y + 2 x + 2 y + 1}{3} )}

= \frac{\frac{2 x + 1}{3} + \frac{2 y + 1}{3}}{1 - ( \frac{2 x ( 2 y + 1 ) + 1 ( 2 y + 1 )}{( 3 ) ^{2}} )}

= \frac{\frac{2 x + 1}{3} + \frac{2 y + 1}{3}}{1 - \frac{( 2 x + 1 )}{3} \frac{( 2 y + 1 )}{3}}

Now,

tan^{- 1} ⎣ ⎡ \frac{\frac{2 y + 1 + 2 x + 1}{3}}{1 - ( \frac{4 x y + 2 x + 2 y + 1}{3} )} ⎦ ⎤ = k

tan^{- 1} tan ⎣ ⎡ \frac{\frac{2 x + 1}{3} + \frac{2 y + 1}{3}}{1 - \frac{( 2 x + 1 )}{3} \frac{( 2 y + 1 )}{3}} ⎦ ⎤ = k

\Rightarrow tan^{- 1} tan [\frac{2 x + 1}{3} + \frac{2 y + 1}{3}] = k

\Rightarrow 2 x + 1 + 2 y + 1 = k 3

\Rightarrow 2 x + 2 y = k 3 - 2

or

x + y = \frac{k 3 - 2}{2}

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Question Text	$tan^{- 1} [\frac{\frac{2 y + 1 + 2 x + 1}{3}}{1 - ( \frac{4 x y + 2 x 2 y + 1}{3} )}] = k$
Answer Type	Text solution:1
Upvotes	150