If x=operatorname{cosec}(tan ^{-1}(cos (cot ^ | vedclass

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(B) Given $x = \operatorname{cosec}(	an ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$.Let $\sin^{-1} a = 	heta$,so $\sin 	heta = a$. Then $\sec \

Vedclass · Accepted Answer

$x^2+a^2=3$

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(B) Given $x = \operatorname{cosec}(	an ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$.Let $\sin^{-1} a = 	heta$,so $\sin 	heta = a$. Then $\sec 	heta = \frac{1}{\sqrt{1-a^2}}$.Now,$x = \operatorname{cosec}(	an ^{-1}(\cos (\cot ^{-1}(\frac{1}{\sqrt{1-a^2}}))))$.Let $\cot^{-1}(\frac{1}{\sqrt{1-a^2}}) = \phi$,so $\cot \phi = \frac{1}{\sqrt{1-a^2}}$.Then $\cos \phi = \frac{\cot \phi}{\sqrt{1+\cot^2 \phi}} = \frac{1/\sqrt{1-a^2}}{\sqrt{1 + 1/(1-a^2)}} = \frac{1}{\sqrt{1-a^2+1}} = \frac{1}{\sqrt{2-a^2}}$.So,${x = \operatorname{cosec}(	an ^{-1}(\frac{1}{\sqrt{2-a^2}}}))$.Let $	an^{-1}(\frac{1}{\sqrt{2-a^2}}) = \psi$,so $	an \psi = \frac{1}{\sqrt{2-a^2}}$.Then $\operatorname{cosec} \psi = \sqrt{1+\cot^2 \psi} = \sqrt{1+(2-a^2)} = \sqrt{3-a^2}$.Thus,$x = \sqrt{3-a^2}$,which implies $x^2 = 3-a^2$,or $x^2+a^2=3$.

If $x=\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$,where $a \in [0, 1]$,then which of the following is true?

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