# If $x= cosec \lfloor\tan^{-1}\{\cos(\cot^{-1}(\sec(\sin^{-1}a)))\}\rfloor\;$ and $y=\sec\lfloor\cot^{-1}\{\sin(\tan^{-1}(cosec(\cos^{-1}a)))\}\rfloor\;$ where $\;a\;\in [0,1]$ Find the relationship between $x$ and $y$ in terms of $a$

Toolbox:
• $$sin^{-1}a=sec^{-1}\large\frac{1}{\sqrt{1-a^2}}$$
• $$cos^{-1}a=cosec^{-1}\large\frac{1}{\sqrt{1-a^2}}$$
• $$cot^{-1}\large\frac{1}{\sqrt{1-a^2}}=cos^{-1}\large\frac{1}{\sqrt{2-a^2}}$$
• $$tan^{-1}\large\frac{1}{\sqrt{1-a^2}}=sin^{-1}\large\frac{1}{\sqrt{2-a^2}}$$
$$put\:sin^{-1}a=\alpha\:\Rightarrow\:sin\alpha=a$$

We know that $$cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-a^2}$$
$$and\:sec\alpha=\large\frac{1}{cos\alpha}=\large\frac{1}{\sqrt{1-a^2}}$$
$$sec\:sin^{-1}a=sec\alpha=\large\frac{1}{\sqrt{1-a^2}}$$
$$\Rightarrow\:cot^{-1}sec(sin^{-1}a)=cot^{-1}\large\frac{1}{\sqrt{1-a^2}}$$

Put $$cot^{-1}\large\frac{1}{\sqrt{1-a^2}}= \beta$$
$$\Rightarrow\:cot\beta=\large\frac{1}{\sqrt{1-a^2}}$$
In any rt angled triangle $$cot\beta=\large\frac{base}{perpendicular}\:and\:cos\beta=\large\frac{base}{hypotenuse}$$
$$(base)^2+(perpendicular)^2=(hypotenuse)^2$$

Substituting the values of base and perpendicular from $$cot\beta$$ we get
$$hypotenuse=\sqrt{1+1-a^2}=\sqrt{2-a^2}$$ and hence
$$\Rightarrow\:cos\beta=\large\frac{1}{\sqrt{2-a^2}}$$
$$cos(cot^{-1}\large\frac{1}{\sqrt{1-a^2}})=cos\beta$$
$$=\large\frac{1}{\sqrt{2-a^2}}$$

Substituting the values in x we get
$$x = cosec \bigg[ tan^{-1} \{ cos \bigg( cot^{-1} \bigg( sec \bigg( sin^{-1}a \bigg)\bigg) \bigg) \} \bigg]$$
$$= cosec \bigg[ tan^{-1} \{ cos \bigg( cot^{-1}\large \frac{1}{\sqrt{1-a^2}} \bigg) \} \bigg]$$
$$cosec \bigg[ \{ tan^{-1} \bigg( \frac{1}{\sqrt{2-a^2}} \bigg) \} \bigg]$$
$$= cosec \: tan^{-1} \large\frac{1}{\sqrt{2-a^2}}$$

Take $$tan^{-1}\large \frac{1}{\sqrt{2-a^2}}=\gamma$$
$$\Rightarrow\:\tan\gamma=\large\frac{1}{\sqrt{2-a^2}}$$
$$\Rightarrow\:x=cosec\gamma=\sqrt{3-a^2}$$
In the same manner using the above procedure we can prove
Similarly $$y = \sqrt{3-a^2}$$
$$\Rightarrow x^2=y^2=3-a^2$$

answered Feb 19, 2013
edited Mar 19, 2013