value of tan−1⁡sin⁡2−1cos⁡2 is

A
$\frac{π}{2} - 1$
B
$1 - \frac{π}{4}$
C
$2 - \frac{π}{2}$
D
$\frac{π}{4} - 1$

Solution:

$\begin{matrix} \frac{\sin 2 - 1}{\cos 2} = - \frac{1 - \sin 2}{\cos 2} = - \frac{(\cos 1 - \sin 1)^{2}}{(\cos 1 + \sin 1) (\cos 1 - \sin 1)} \\ = - \frac{\cos 1 - \sin 1}{\cos 1 + \sin 1} = - \frac{1 - \tan 1}{1 + \tan 1} \\ = - \tan (\frac{π}{4} - 1) = \tan (1 - \frac{π}{4}) \\ \Rightarrow \tan^{- 1} (\frac{\sin 2 - 1}{\cos 2}) = \tan^{- 1} [\tan (1 - \frac{π}{4})] = 1 - \frac{π}{4} \end{matrix}$

value of $\tan^{- 1} (\frac{\sin 2 - 1}{\cos 2})$ is

Solution:

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