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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 $a$

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  • \( sin^{-1}a=sec^{-1}\large\frac{1}{\sqrt{1-a^2}}\)
  • \( cos^{-1}a=cosec^{-1}\large\frac{1}{\sqrt{1-a^2}}\)
  • \( cot^{-1}\large\frac{1}{\sqrt{1-a^2}}=cos^{-1}\large\frac{1}{\sqrt{2-a^2}}\)
  • \( tan^{-1}\large\frac{1}{\sqrt{1-a^2}}=sin^{-1}\large\frac{1}{\sqrt{2-a^2}}\)
We know that \(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-a^2}\)
Put \(cot^{-1}\large\frac{1}{\sqrt{1-a^2}}= \beta\)
In any rt angled triangle \(cot\beta=\large\frac{base}{perpendicular}\:and\:cos\beta=\large\frac{base}{hypotenuse}\)
Substituting the values of base and perpendicular from \( cot\beta\) we get
\(hypotenuse=\sqrt{1+1-a^2}=\sqrt{2-a^2}\) and hence
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 \(tan^{-1}\large \frac{1}{\sqrt{2-a^2}}=\gamma\)
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 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1

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