# Sin(sin^-1x+cos^-1x) find the value?

$\sin(\sin^{-1}x+\cos^{-1}x)\\* using\;\sin(A-B)=\sin(A) \cos(B)-\sin(B) \cos(A) \\* Now,\;we\;have,\\* \Rightarrow \sin(\sin^{-1}x-\cos^{-1}x)=\sin(\sin^{-1}x) \cos( \cos^{-1}x)-\sin(\cos^{-1}x) \cos( \sin^{-1}x) \\* \sin(\sin^{-1}x)\;and \;\cos(\cos^{-1}x)\;both\;equal\;x\;because\;you\;are\;taking\;inverse \; functions.\\*So,\;it\;will\; reduced\;to,\\* x\times x-\sin(\cos^{-1}x)\cos(\sin^{-1}x)\\*Let\;A=\sin(\cos^{-1}x),\;Let\;\theta = \cos^{-1}x.\\* Then, \;substituting,\; we \;get\; A=sin(\theta).\\* Let\;look\; at\; \theta =\cos^{-1}x\;for\;a\;bit.\;It\;means\;x=\cos(\theta)\\* But \;if \;A= \sin(\theta) \;and \;x=\cos(\theta),\;and\;since\;\sin^2(\theta)+\cos^2(\theta) =1, \\*that \;means \;A^2 + x^2 =1 \\*Thus\;A=\sqrt{(1-x^2)} \\*Now,\;it \;turns \;out \;that\; both\; \cos(\sin^{-1}x) \;and\; \sin( \cos^{-1}x ) \;equal\; \sqrt{(1-x^2)}\\* \Rightarrow x\times x-\sqrt{1-x^2}\times \sqrt{1-x^2}\\* \Rightarrow x^2-(1-x^2)=2x^2-1\\$

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