The value of tan^{-1} left( frac{sin 2 - 1}{c | vedclass

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(C) We know that $\sin 2 = 2 \sin 1 \cos 1$ and $\cos 2 = \cos^2 1 - \sin^2 1$. Also,$1 = \sin^2 1 + \cos^2 1$.Substituting these in the expression:$\

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$1 - \frac{\pi}{4}$

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(C) We know that $\sin 2 = 2 \sin 1 \cos 1$ and $\cos 2 = \cos^2 1 - \sin^2 1$. Also,$1 = \sin^2 1 + \cos^2 1$.Substituting these in the expression:$	an^{-1} \left( \frac{2 \sin 1 \cos 1 - (\sin^2 1 + \cos^2 1)}{\cos^2 1 - \sin^2 1} ight)$$= 	an^{-1} \left( \frac{-( \cos^2 1 - 2 \sin 1 \cos 1 + \sin^2 1)}{\cos^2 1 - \sin^2 1} ight)$$= 	an^{-1} \left( \frac{-(\cos 1 - \sin 1)^2}{(\cos 1 - \sin 1)(\cos 1 + \sin 1)} ight)$$= 	an^{-1} \left( -\frac{\cos 1 - \sin 1}{\cos 1 + \sin 1} ight)$$= 	an^{-1} \left( \frac{\sin 1 - \cos 1}{\cos 1 + \sin 1} ight)$Dividing numerator and denominator by $\cos 1$:$= 	an^{-1} \left( \frac{	an 1 - 1}{1 + 	an 1} ight)$$= 	an^{-1} \left( 	an(1 - \frac{\pi}{4}) ight)$$= 1 - \frac{\pi}{4}$

The value of $\tan^{-1} \left( \frac{\sin 2 - 1}{\cos 2} \right)$ is equal to:

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