Tan -1 - 2y-1-2x-1 - 3 - 1- 4xy-2x2y-1 - 3 - -k

Consider 2y+1+2x+1√(3)1 (4xy+2x+2y+13) =2x+1√(3)+2y+1√(3)1 (2x(2y+1)+1(2y+1)(√(3))^2) =2x+1√(3)+2y+1√(3)1 (2x+1)√(3)(2y+1)√(3) Now, tan^ 1[2 ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

tan -1 [ 2y+1...

Question

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

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Solution

Consider $\frac{\frac{2 y + 1 + 2 x + 1}{\sqrt{3}}}{1 - (\frac{4 x y + 2 x + 2 y + 1}{3})}$
$= \frac{\frac{2 x + 1}{\sqrt{3}} + \frac{2 y + 1}{\sqrt{3}}}{1 - ⎛ ⎜ ⎜ ⎝ \frac{2 x (2 y + 1) + 1 (2 y + 1)}{{(\sqrt{3})}^{2}} ⎞ ⎟ ⎟ ⎠}$
$= \frac{\frac{2 x + 1}{\sqrt{3}} + \frac{2 y + 1}{\sqrt{3}}}{1 - \frac{(2 x + 1)}{\sqrt{3}} \frac{(2 y + 1)}{\sqrt{3}}}$
Now, ${tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{\frac{2 y + 1 + 2 x + 1}{\sqrt{3}}}{1 - (\frac{4 x y + 2 x + 2 y + 1}{3})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = k$
${tan}^{- 1} tan ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{\frac{2 x + 1}{\sqrt{3}} + \frac{2 y + 1}{\sqrt{3}}}{1 - \frac{(2 x + 1)}{\sqrt{3}} \frac{(2 y + 1)}{\sqrt{3}}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = k$
$\Rightarrow {tan}^{- 1} tan [\frac{2 x + 1}{\sqrt{3}} + \frac{2 y + 1}{\sqrt{3}}] = k$
$\Rightarrow 2 x + 1 + 2 y + 1 = k \sqrt{3}$
$\Rightarrow 2 x + 2 y = k \sqrt{3} - 2$
or $x + y = \frac{k \sqrt{3} - 2}{2}$

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Similar questions

Q. If

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

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tan−1⎡⎣2y+1+2x+1√31−(4xy+2x2y+13)⎤⎦=k

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