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

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(B) Given expression: $	an ^{-1}\left(\frac{\cos \left(\frac{15 \pi}{4}ight)-1}{\sin \left(\frac{\pi}{4}ight)}ight)$ Since $\frac{15 \pi}{4} =

Vedclass · Accepted Answer

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

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(B) Given expression: $	an ^{-1}\left(\frac{\cos \left(\frac{15 \pi}{4}ight)-1}{\sin \left(\frac{\pi}{4}ight)}ight)$ Since $\frac{15 \pi}{4} = 4 \pi - \frac{\pi}{4}$,we have $\cos \left(\frac{15 \pi}{4}ight) = \cos \left(4 \pi - \frac{\pi}{4}ight) = \cos \left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$. Substituting this into the expression: $	an ^{-1}\left(\frac{\frac{1}{\sqrt{2}}-1}{\frac{1}{\sqrt{2}}}ight) = 	an ^{-1}\left(\frac{1-\sqrt{2}}{1}ight) = 	an ^{-1}(1-\sqrt{2})$. Using the identity $	an \left(\frac{	heta}{2}ight) = \frac{1-\cos 	heta}{\sin 	heta}$,we have $	an \left(\frac{\pi}{8}ight) = \sqrt{2}-1$. Therefore,$	an \left(-\frac{\pi}{8}ight) = -(\sqrt{2}-1) = 1-\sqrt{2}$. Thus,$	an ^{-1}(1-\sqrt{2}) = -\frac{\pi}{8}$.

The value of $\tan ^{-1}\left(\frac{\cos \left(\frac{15 \pi}{4}\right)-1}{\sin \left(\frac{\pi}{4}\right)}\right)$ is equal to

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