If sin ^{ - 1}frac{{2a}}{{1 + {a^2}}} - cos ^ | vedclass

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(D) Let $a = 	an 	heta$,$b = 	an \phi$,and $x = 	an \psi$. Substituting these into the given equation: $\sin ^{ - 1}(\sin 2	heta) - \cos ^{ - 1}(

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$\frac{{a - b}}{{1 + ab}}$

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(D) Let $a = 	an 	heta$,$b = 	an \phi$,and $x = 	an \psi$. Substituting these into the given equation: $\sin ^{ - 1}(\sin 2	heta) - \cos ^{ - 1}(\cos 2\phi) = 	an ^{ - 1}(	an 2\psi)$ $2	heta - 2\phi = 2\psi$ $	heta - \phi = \psi$ Taking the tangent of both sides: $	an(	heta - \phi) = 	an \psi$ Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$: $\frac{	an 	heta - 	an \phi}{1 + 	an 	heta 	an \phi} = 	an \psi$ Substituting back $a, b, x$: $\frac{a - b}{1 + ab} = x$

If $\sin ^{ - 1}\frac{{2a}}{{1 + {a^2}}} - \cos ^{ - 1}\frac{{1 - {b^2}}}{{1 + {b^2}}} = \tan ^{ - 1}\frac{{2x}}{{1 - {x^2}}}$,then $x = $

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