Question

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

• A. $x^2 = y^2 = a^2$
• B. $x^2 = y^2 = 3 + a^2$
• C. $x^2 = y^2 = 3 - a^2$
• D. none of these

Answer 1

Answer: (C)

We have,
$x = cosec\left[tan^{-1}\left\{cos\left(cot^{-1}\left(sec\left(sin^{-1}\,a\right)\right)\right)\right\}\right]$
$= cosec\left[tan^{-1}\left\{cos\left(cot^{-1}\left(sec\left(sec^{-1} \frac{1}{\sqrt{1-a^{2}}}\right)\right)\right)\right\}\right]$
$= cosec\left[tan^{-1}\left\{cos\left(cot^{-1} \frac{1}{\sqrt{1-a^{2}}}\right)\right\}\right]$
$= cosec\left[tan^{-1}\left\{cos\left(cos^{-1} \frac{1}{\sqrt{2-a^{2}}}\right)\right\}\right]$
$= cosec\left(tan^{-1} \frac{1}{\sqrt{2-a^{2}}}\right)$
$= cosec\left( cosec^{-1} \sqrt{3-a^{2}}\right) = \sqrt{3-a^{2}}$
and, $y = sec \left[cot^{-1}\left\{sin\left(tan^{-1}\left(cosec\left(cos^{-1}\,a\right)\right)\right)\right\}\right]$
$= sec \left[cot^{-1}\left\{sin\left(tan^{-1}\left(cosec\left(cosec^{-1}\frac{1}{\sqrt{1-a^{2}}}\right)\right)\right)\right\}\right]$
$= sec \left[cot^{-1}\left\{sin\left(tan^{-1}\frac{1}{\sqrt{1-a^{2}}}\right)\right\}\right]$
$= sec \left[cot^{-1}\left\{sin\left(tan^{-1}\frac{1}{\sqrt{2-a^{2}}}\right)\right\}\right]$
$= sec \left(cot^{-1}\frac{1}{\sqrt{2-a^{2}}}\right)$
$= sec\left( sec^{-1}\sqrt{3-a^{2}}\right) = \sqrt{3-a^{2}}$
$\therefore x^{2} = y^{2} = 3 - a^{2}$