The value of tan^{-1} left( frac{cos(frac{19p | vedclass

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(C) Step $1$: Simplify $\cos(\frac{19\pi}{4})$. Since $\frac{19\pi}{4} = 4\pi + \frac{3\pi}{4}$, we have $\cos(\frac{19\pi}{4}) = \cos(\frac{3\pi}{4})

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$-\frac{3\pi}{8}$

Vedclass · Answer

(C) Step $1$: Simplify $\cos(\frac{19\pi}{4})$. Since $\frac{19\pi}{4} = 4\pi + \frac{3\pi}{4}$, we have $\cos(\frac{19\pi}{4}) = \cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$.Step $2$: Substitute the values into the expression: $\frac{-\frac{1}{\sqrt{2}} - 1}{\frac{1}{\sqrt{2}}} = \frac{-\frac{1+\sqrt{2}}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = -(1+\sqrt{2})$.Step $3$: We need to find $	an^{-1}(-(1+\sqrt{2}))$. Since $	an(\frac{3\pi}{8}) = 	an(67.5^\circ) = 1+\sqrt{2}$, then $	an(-\frac{3\pi}{8}) = -(1+\sqrt{2})$.Step $4$: Therefore, $	an^{-1}(-(1+\sqrt{2})) = -\frac{3\pi}{8}$.

The value of $\tan^{-1} \left( \frac{\cos(\frac{19\pi}{4}) - 1}{\sin(\frac{\pi}{4})} \right)$ is equal to

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