The value of frac{sin 70^{circ}+cos 40^{circ} | vedclass

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Question

(B) We have the expression $\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}$.Using the identity $\cos 	heta = \sin(90^{\circ}

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(B) We have the expression $\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}$.Using the identity $\cos 	heta = \sin(90^{\circ}-	heta)$ and $\sin 	heta = \cos(90^{\circ}-	heta)$:$\frac{\sin 70^{\circ}+\sin 50^{\circ}}{\cos 70^{\circ}+\cos 50^{\circ}}$Using sum-to-product formulas $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$ and $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:$= \frac{2 \sin(\frac{70^{\circ}+50^{\circ}}{2}) \cos(\frac{70^{\circ}-50^{\circ}}{2})}{2 \cos(\frac{70^{\circ}+50^{\circ}}{2}) \cos(\frac{70^{\circ}-50^{\circ}}{2})}$$= \frac{2 \sin 60^{\circ} \cos 10^{\circ}}{2 \cos 60^{\circ} \cos 10^{\circ}}$$= \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = 	an 60^{\circ} = \sqrt{3}$.

The value of $\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}$ is equal to

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