If $x=\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\sec \left(\sin ^{-1}…

If $x=\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\sec \left(\sin ^{-1} a\right)\right)\right)\right)\right)$, $\mathrm{a} \in[0,1]$
  1. $x^2-\mathrm{a}^2=3$
  2. $x^2+\mathrm{a}^2=3$
  3. $x^2-\mathrm{a}^2=2$
  4. $x^2+\mathrm{a}^2=2$

Solution

$\begin{aligned} & x=\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\sec \left(\sin ^{-1} \mathrm{a}\right)\right)\right)\right)\right) \\ & =\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\sec \left(\sec ^{-1} \frac{1}{\sqrt{1-\mathrm{a}^2}}\right)\right)\right)\right)\right) \\ & =\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\frac{1}{\sqrt{1-\mathrm{a}^2}}\right)\right)\right)\right) \\ & =\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cos ^{-1} \frac{1}{\sqrt{2-\mathrm{a}^2}}\right)\right)\right) \\ & =\operatorname{cosec}\left(\tan ^{-1}\left(\frac{1}{\sqrt{2-\mathrm{a}^2}}\right)\right) \\ & =\operatorname{cosec}\left(\operatorname{cosec}^{-1}\left(\sqrt{3-\mathrm{a}^2}\right)\right) \\ \therefore \quad & x=\sqrt{3-\mathrm{a}^2} \\ \therefore \quad & x^2+\mathrm{a}^2=3\end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 1)

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