The value of cos 105^circ + sin 105^circ is | vedclass

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(D) We know that $\cos 105^\circ = \cos(60^\circ + 45^\circ) = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ = \frac{1}{2} \cdot \frac{1}{

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$\frac{1}{\sqrt{2}}$

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(D) We know that $\cos 105^\circ = \cos(60^\circ + 45^\circ) = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$.Similarly,$\sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.Adding these two values:$\cos 105^\circ + \sin 105^\circ = \frac{1 - \sqrt{3}}{2\sqrt{2}} + \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{1 - \sqrt{3} + \sqrt{3} + 1}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

The value of $\cos 105^\circ + \sin 105^\circ$ is

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