frac{sqrt{2} cos 45^{circ}+cos 56^{circ}+cos | vedclass

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Question

(A) Let $A = \frac{\sqrt{2} \cos 45^{\circ} + \cos 56^{\circ} + \cos 58^{\circ} - \cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) Let $A = \frac{\sqrt{2} \cos 45^{\circ} + \cos 56^{\circ} + \cos 58^{\circ} - \cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the numerator becomes $1 + \cos 56^{\circ} + \cos 58^{\circ} - \cos 66^{\circ}$.Using $\cos 58^{\circ} - \cos 66^{\circ} = 2 \sin 62^{\circ} \sin 4^{\circ}$ is not helpful,so we rearrange: $1 - \cos 66^{\circ} + \cos 56^{\circ} + \cos 58^{\circ} = 2 \sin^2 33^{\circ} + 2 \cos 57^{\circ} \cos 1^{\circ}$.Since $\sin 33^{\circ} = \cos 57^{\circ}$,the numerator is $2 \cos 57^{\circ} (\cos 57^{\circ} + \cos 1^{\circ})$.The denominator is $\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \cos 57^{\circ}$.Using $2 \cos 28^{\circ} \cos 29^{\circ} = \cos 57^{\circ} + \cos 1^{\circ}$,the expression simplifies to $\frac{2 \cos 57^{\circ} (\cos 57^{\circ} + \cos 1^{\circ})}{\sqrt{2} \cos 57^{\circ} (\cos 57^{\circ} + \cos 1^{\circ})} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

List-$I$	List-$II$
$(I)$ $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \dots + \sin^2 90^{\circ}$	$(A)$ $0$
$(II)$ $\tan^2 5^{\circ} \cdot \tan^2 10^{\circ} \cdot \tan^2 15^{\circ} \dots \tan^2 85^{\circ}$	$(B)$ $\frac{19}{2}$
$(III)$ $\cos^2 5^{\circ} + \cos^2 10^{\circ} + \cos^2 15^{\circ} + \dots + \cos^2 180^{\circ}$	$(C)$ $18$
$(IV)$ $\cot 5^{\circ} + \cot 10^{\circ} + \cot 15^{\circ} + \dots + \cot 175^{\circ}$	$(D)$ $1$
	$(E)$ $-1$

$\frac{\sqrt{2} \cos 45^{\circ}+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}} = ?$

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