Hi, sincot-1tancos-1xwe know that cos-1x=tan-1 1-x2xtancos-1x=tantan-1 1-x2x= 1-x2x=sincot-1tancos-1x=sincot-11-x2xlet cot-11-x2x= t so we have to find sint cot t = 1-x2x=baseperpendicularhypotenuse = x2+1-x2=1so sin t = x1= x = RHS Regards

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Hi, \sin [[\cot^{- 1} \{\tan (\cos^{- 1} x)\}]] we know that \cos^{- 1} x = \tan^{- 1} \frac{\sqrt{1 - x^{2}}}{x} \tan (\cos^{- 1} x) = \tan (\tan^{- 1} \frac{\sqrt{1 - x^{2}}}{x}) = \frac{\sqrt{1 - x^{2}}}{x} = \sin [[\cot^{- 1} \{\tan (\cos^{- 1} x)\}]] = \sin [[\cot^{- 1} \{\frac{\sqrt{1 - x^{2}}}{x}\}]] let \cot^{- 1} \frac{\sqrt{1 - x^{2}}}{x} = t so we have to find sint \cot t = \frac{\sqrt{1 - x^{2}}}{x} = \frac{base}{perpendicular} hypotenuse = \sqrt{x^{2} + 1 - x^{2}} = 1 so \sin t = \frac{x}{1} = x = RHS

Regards

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