Question

If $x$ and $y$ are the solutions of the equation $12sin x+5cos ⁡ x=2y^{2}-8y+21$ , then the value of $12cot\left(\frac{x y}{2}\right)$ is (Given, $\left|x\right| < \pi $ )

Answer 1

$LHS=12sin x+5cos ⁡ x\in \left[- \sqrt{1 2^{2} + 5^{2}} , \sqrt{1 2^{2} + 5^{2}}\right]$
i.e. $\left[- 13,13\right]$ i.e.
maximum value of $LHS$ is $13$
$RHS=2\left(y^{2} - 4 y + 4\right)+13$
$=2\left(y - 2\right)^{2}+13$
$RHS\geq 13$
Roots of the equation exist if $LHS=RHS=13.$
$RHS=13$ when $y=2$
$LHS=13\Rightarrow 12sin x+5cos ⁡ x=13$
$\Rightarrow \frac{12}{13}sin x+\frac{5}{13}cos ⁡ x=1$
$sin\left(x + \alpha \right)=1,$ where $tan \alpha =\frac{5}{12}$
$x+tan^{- 1}\frac{5}{12}=\frac{\pi }{2}\Rightarrow x=\frac{\pi }{2}-tan^{- 1}\frac{5}{12}$
$\Rightarrow xy=\pi -2tan^{- 1}\frac{5}{12}\Rightarrow \frac{5}{12}=tan\left(\frac{\pi }{2} - \frac{x y}{2}\right)=cot\left(\frac{x y}{2}\right)$
$\Rightarrow 12cot\left(\frac{x y}{2}\right)=5$