Solve for $x a n d y$
$1 - 2 x - x^{2} = t a n^{2} (x + y) + c o t^{2} (x + y)$

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Solution

Given, $1 - 2 x - x^{2} = {tan}^{2} (x + y) + {cot}^{2} (x + y) \Rightarrow 2 - {(x + 1)}^{2} = {tan}^{2} (x + y) + {cot}^{2} (x + y)$
As $R . H . S \geq 2$ and $L . H S \leq 2$
This is only possible when
$L . H . S = R . H . S = 2 \Rightarrow x = - 1 {tan}^{2} (x + y) = 1 \Rightarrow {tan}^{2} (y - 1) = 1$
$\Rightarrow y - 1 = n π \pm \frac{π}{4} \Rightarrow y = n π \pm \frac{π}{4} + 1$

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Similar questions

1 - 2 x - x^{2} = {tan}^{2} (x + y) + {cot}^{2} (x + y)

{tan}^{2} (x + y) + {cot}^{2} (x + y) = 1 + 2 x - x^{2},

\frac{2}{\sqrt{3}} (t a n x - c o t x) = t a n^{2} x + c o t^{2} x - 2.

{tan}^{2} (x + y) + {cot}^{2} (x + y) = 1 - 2 x - x^{2}

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

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