Inverse Trigonometric Functions

${\text{tan}}^{-1}4x+{\text{tan}}^{-1}6x=\frac{\pi }{6}$

$\Rightarrow {\text{tan}}^{-1}\left(\frac{4x+6x}{1-24x^2}\right)=\frac{\pi }{6}$

$\Rightarrow \frac{10x}{1-24x^2}=\frac{1}{\sqrt{3}}$

$\Rightarrow 24x^2+10\sqrt{3}x-1=0$

$x=\frac{-10\sqrt{3}\pm \sqrt{300+96}}{48}$

$x=\frac{\sqrt{396}-10\sqrt{3}}{48}$

Only 1 solution in $\left(-\frac{1}{2\sqrt{6}},\frac{1}{2\sqrt{6}}\right)$

$$\begin{gathered} f\left(x\right)=16\left({\left({\sec }^{-1}x\right)}^2+{\left({cosec}^{-1}x\right)}^2\right) \\ \Rightarrow f\left(x\right)=16\left({\left({\sec }^{-1}x\right)}^2+{\left(\frac{\mathrm{\pi }}{2}-{\sec }^{-1}x\right)}^2\right) \\ \Rightarrow f\left(x\right)=32\left({\left({\sec }^{-1}x-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{16}\right) \\ \Rightarrow f{\left(x\right)}_{min}=32\times \frac{{\mathrm{\pi }}^2}{16}=2{\mathrm{\pi }}^2 \\ \Rightarrow \mathrm{f}{\left(\mathrm{x}\right)}_{\max }=32\left({\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{16}\right)=20{\mathrm{\pi }}^2 \\ \Rightarrow \mathrm{f}{\left(\mathrm{x}\right)}_{\min }+\mathrm{f}{\left(\mathrm{x}\right)}_{\max }=22{\mathrm{\pi }}^2 \end{gathered}$$

Given \(y=\sin ^{-1} x,-1 \leq x \leq 0\)

\(\sin ^{-1} x\) is increasing on \([-1,1]\)

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

\(x=0 \Rightarrow y=\sin ^{-1}(0)=0\)

⇒ Range of \(y=\left[-\frac{\pi}{2}, 0\right]\)

Given \(y=\sin ^{-1} x,-1 \leq x \leq 0\)

\(\sin ^{-1} x\) is increasing on \([-1,1]\)

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

\(x=0 \Rightarrow y=\sin ^{-1}(0)=0\)

$\Rightarrow$ Range of \(y=\left[-\frac{\pi}{2}, 0\right]\)