The value of frac{sin 55^{circ} - cos 55^{cir | vedclass

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(D) We know that $\cos 55^{\circ} = \sin(90^{\circ} - 55^{\circ}) = \sin 35^{\circ}$.Substituting this into the expression,we get $\frac{\sin 55^{\cir

Vedclass · Accepted Answer

$\sqrt{2}$

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(D) We know that $\cos 55^{\circ} = \sin(90^{\circ} - 55^{\circ}) = \sin 35^{\circ}$.Substituting this into the expression,we get $\frac{\sin 55^{\circ} - \sin 35^{\circ}}{\sin 10^{\circ}}$.Using the formula $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}ight) \sin\left(\frac{C-D}{2}ight)$,we have:$\sin 55^{\circ} - \sin 35^{\circ} = 2 \cos\left(\frac{55^{\circ} + 35^{\circ}}{2}ight) \sin\left(\frac{55^{\circ} - 35^{\circ}}{2}ight) = 2 \cos 45^{\circ} \sin 10^{\circ}$.Now,the expression becomes $\frac{2 \cos 45^{\circ} \sin 10^{\circ}}{\sin 10^{\circ}} = 2 \cos 45^{\circ}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the value is $2 	imes \frac{1}{\sqrt{2}} = \sqrt{2}$.

The value of $\frac{\sin 55^{\circ} - \cos 55^{\circ}}{\sin 10^{\circ}}$ is

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