logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

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 $a$

Can you answer this question?
 
 

1 Answer

0 votes
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 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...