Question

For $k \in R$, let the solutions of the equation $\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k, 0<|x|<\frac{1}{\sqrt{2}}$ be $\alpha$ and $\beta$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $x ^2- bx -5=0$ are $\frac{1}{\alpha^2}+\frac{1}{\beta^2}$ and $\frac{\alpha}{\beta}$, then $\frac{b}{k^2}$ is equal to ______

Answer 1

$ \cos \left(\sin ^{-1} x\right)=\cos \left(\cos ^{-1} \sqrt{1-x^2}\right)=\sqrt{1-x^2} $
$ \cot \left(\tan ^{-1} \sqrt{1-x^2}\right)=\cot \cot ^{-1}\left(\sqrt{\frac{1}{\sqrt{1-x^2}}}\right)=\frac{1}{\sqrt{1-x^2}} $
$ \Rightarrow \cos \left(\sin ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right)=\frac{\sqrt{1-2 x^2}}{\sqrt{1-x^2}} $
$ \Rightarrow \frac{\sqrt{1-2 x^2}}{\sqrt{1-x^2}}=k$
$ \Rightarrow 1-2 x^2=k^2\left(1-x^2\right)$
$ \Rightarrow\left(k^2-2\right) x^2=k^2-1$
$ x^2=\frac{k^2-1}{k^2-2} $
$ \alpha=\sqrt{\frac{k^2-1}{k^2-2}} \Rightarrow \alpha^2=\frac{k^2-1}{k^2-2}$
$ \beta=\sqrt{\frac{k^2-1}{k^2-2}} \Rightarrow \beta^2=\frac{k^2-1}{k^2-2}$
$ \frac{1}{\alpha^2}+\frac{1}{\beta^2}=2\left(\frac{k^2-2}{k^2-1}\right) \& \frac{\alpha}{\beta}=-1 $
Sum of roots $=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{\alpha}{\beta}=b $
$ \Rightarrow \frac{2\left(k^2-2\right)}{k^2-1}-1=b \ldots .(1) $
Product of roots $=\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right) \frac{\alpha}{\beta}=-5$
$ \Rightarrow \frac{2\left(k^2-2\right)}{k^2-1}(-1)=-5 $
$ \Rightarrow 2 k^2-4=5 k^2-5$
$ \Rightarrow 3 k^2=1 \Rightarrow k^2=\frac{1}{3} \ldots . \text { Put in (1) }$
$ \Rightarrow b=\frac{2\left(k^2-2\right)}{k^2-1}-1=5-1=4 $
$ \frac{b}{k^2}=\frac{4}{\frac{1}{3}}=12$