Question

If x =cosec(tan inverse (cos(cot inverse(sec(sin inverse a))
Y=sec(cotinverse(sin(tan inverse(cosec(cos inverse a))
Find the relationship between x and y in terms of a

Answer 1

$$\begin{gathered} \mathrm{If}\mathrm{x}=\mathrm{cosec}\left({\tan }^{-1}\left(\cos \left(\cot {}^{-1}\left(\sec \left({\sin }^{-1}\mathrm{a}\right)\right)\right)\right)\right) \\ \mathrm{y}=\sec \left({\cot }^{-1}\left(\sin \left(\tan {}^{-1}\left(\mathrm{cosec}\left(\cos {}^{-1}\mathrm{a}\right)\right)\right)\right)\right) \\ \mathrm{solving}\mathrm{for}\mathrm{x} \\ \mathrm{x}=\mathrm{cosec}\left({\tan }^{-1}\left(\cos \left(\cot {}^{-1}\left(\sec \left({\sin }^{-1}\mathrm{a}\right)\right)\right)\right)\right) \\ =\mathrm{cosec}\left({\tan }^{-1}\left(\cos \left(\cot {}^{-1}\left(\frac{1}{\cos \left({\sin }^{-1}\mathrm{a}\right)}\right)\right)\right)\right) \\ =\mathrm{cosec}\left({\tan }^{-1}\left(\cos \left(\cot {}^{-1}\left(\frac{1}{\cos ({\cos }^{-1}\sqrt{1-{\mathrm{a}}^2}}\right)\right)\right)\right) \\ =\mathrm{cosec}\left({\tan }^{-1}\left(\cos \left(\cot {}^{-1}\left(\frac{1}{\sqrt{1-{\mathrm{a}}^2}}\right)\right)\right)\right) \\ =\mathrm{cosec}\left({\tan }^{-1}\left(\cos \left(\cos {}^{-1}\left(\frac{1}{\sqrt{{\left(\sqrt{1-{\mathrm{a}}^2}\right)}^2+1}}\right)\right)\right)\right) \\ =\mathrm{cosec}\left({\tan }^{-1}\left(\cos \left(\cos {}^{-1}\left(\frac{1}{\sqrt{2-{\mathrm{a}}^2}}\right)\right)\right)\right) \\ =\mathrm{cosec}\left({\tan }^{-1}\left(\frac{1}{\sqrt{2-{\mathrm{a}}^2}}\right)\right) \\ =\frac{1}{\sin \left({\tan }^{-1}\left(\frac{1}{\sqrt{2-{\mathrm{a}}^2}}\right)\right)} \\ =\frac{1}{\sin \left({\sin }^{-1}\left({\displaystyle \frac{1}{\sqrt{{\left(\sqrt{2-{\mathrm{a}}^2}\right)}^2+1}}}\right)\right)} \\ =\frac{1}{\frac{1}{\sqrt{3-{\mathrm{a}}^2}}} \\ \mathrm{x}=\sqrt{3-{\mathrm{a}}^2}\dots \left(1\right) \\ \mathrm{solving}\mathrm{for}\mathrm{y} \\ \mathrm{y}=\sec \left({\cot }^{-1}\left(\sin \left(\tan {}^{-1}\left(\mathrm{cosec}\left(\cos {}^{-1}\mathrm{a}\right)\right)\right)\right)\right) \\ =\sec \left({\cot }^{-1}\left(\sin \left(\tan {}^{-1}\left(\frac{1}{\sin \left(\cos {}^{-1}\mathrm{a}\right)}\right)\right)\right)\right) \\ =\sec \left({\cot }^{-1}\left(\sin \left(\tan {}^{-1}\left(\frac{1}{\sin \left(\sin {}^{-1}\sqrt{1-{\mathrm{a}}^2}\right)}\right)\right)\right)\right) \\ =\sec \left({\cot }^{-1}\left(\sin \left(\tan {}^{-1}\left(\frac{1}{\sqrt{1-{\mathrm{a}}^2}}\right)\right)\right)\right) \\ =\sec \left({\cot }^{-1}\left(\sin \left(\sin {}^{-1}\left(\frac{1}{\sqrt{{\left(\sqrt{1-{\mathrm{a}}^2}\right)}^2+1}}\right)\right)\right)\right) \\ =\sec \left({\cot }^{-1}\left(\frac{1}{\sqrt{{\left(\sqrt{1-{\mathrm{a}}^2}\right)}^2+1}}\right)\right) \\ =\sec \left({\cot }^{-1}\left(\frac{1}{\sqrt{2-{\mathrm{a}}^2}}\right)\right) \\ =\frac{1}{\cos \left({\cos }^{-1}\left({\displaystyle \frac{1}{\sqrt{{\left(\sqrt{2-{\mathrm{a}}^2}\right)}^2+1}}}\right)\right)} \\ =\frac{1}{\cos \left({\cos }^{-1}\left({\displaystyle \frac{1}{\sqrt{3-{\mathrm{a}}^2}}}\right)\right)} \\ =\frac{1}{{\displaystyle \frac{1}{\sqrt{3-{\mathrm{a}}^2}}}} \\ \mathrm{y}=\sqrt{3-{\mathrm{a}}^2}\dots \left(2\right) \\ \mathrm{so},\mathrm{from}\mathrm{eq}\left(1\right)\&\hspace{0.17em}\left(2\right) \\ \mathrm{x}=\mathrm{y}=\sqrt{3-{\mathrm{a}}^2} \end{gathered}$$