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

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(B) આપેલ છે કે $x = \operatorname{cosec}(	an ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$.ધારો કે $\sin^{-1} a = 	heta$,તેથી $\sin 	heta = a$. ત

Vedclass · Accepted Answer

$x^2+a^2=3$

Vedclass · Answer

(B) આપેલ છે કે $x = \operatorname{cosec}(	an ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$.ધારો કે $\sin^{-1} a = 	heta$,તેથી $\sin 	heta = a$. તો $\sec 	heta = \frac{1}{\sqrt{1-a^2}}$.હવે,$x = \operatorname{cosec}(	an ^{-1}(\cos (\cot ^{-1}(\frac{1}{\sqrt{1-a^2}}))))$.ધારો કે $\cot^{-1}(\frac{1}{\sqrt{1-a^2}}) = \phi$,તેથી $\cot \phi = \frac{1}{\sqrt{1-a^2}}$.તો $\cos \phi = \frac{\cot \phi}{\sqrt{1+\cot^2 \phi}} = \frac{1/\sqrt{1-a^2}}{\sqrt{1 + 1/(1-a^2)}} = \frac{1}{\sqrt{1-a^2+1}} = \frac{1}{\sqrt{2-a^2}}$.તેથી,${x = \operatorname{cosec}(	an ^{-1}(\frac{1}{\sqrt{2-a^2}}}))$.ધારો કે $	an^{-1}(\frac{1}{\sqrt{2-a^2}}) = \psi$,તેથી $	an \psi = \frac{1}{\sqrt{2-a^2}}$.તો $\operatorname{cosec} \psi = \sqrt{1+\cot^2 \psi} = \sqrt{1+(2-a^2)} = \sqrt{3-a^2}$.આમ,$x = \sqrt{3-a^2}$,જે સૂચવે છે કે $x^2 = 3-a^2$,અથવા $x^2+a^2=3$.

જો $x=\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$,જ્યાં $a \in [0, 1]$,તો નીચેનામાંથી કયું સાચું છે?

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