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

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Question

(C) Given expression: $\frac{\sin 70^\circ + \cos 40^\circ}{\cos 70^\circ + \sin 40^\circ}$Using the identities $\cos 	heta = \sin(90^\circ - 	heta)

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(C) Given expression: $\frac{\sin 70^\circ + \cos 40^\circ}{\cos 70^\circ + \sin 40^\circ}$Using the identities $\cos 	heta = \sin(90^\circ - 	heta)$:$\cos 40^\circ = \sin(90^\circ - 40^\circ) = \sin 50^\circ$$\cos 70^\circ = \sin(90^\circ - 70^\circ) = \sin 20^\circ$Substituting these values:$= \frac{\sin 70^\circ + \sin 50^\circ}{\sin 20^\circ + \sin 40^\circ}$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:Numerator: $\sin 70^\circ + \sin 50^\circ = 2 \sin(\frac{70^\circ + 50^\circ}{2}) \cos(\frac{70^\circ - 50^\circ}{2}) = 2 \sin 60^\circ \cos 10^\circ$Denominator: $\sin 20^\circ + \sin 40^\circ = 2 \sin(\frac{20^\circ + 40^\circ}{2}) \cos(\frac{20^\circ - 40^\circ}{2}) = 2 \sin 30^\circ \cos(-10^\circ)$Since $\cos(-10^\circ) = \cos 10^\circ$:$= \frac{2 \sin 60^\circ \cos 10^\circ}{2 \sin 30^\circ \cos 10^\circ} = \frac{\sin 60^\circ}{\sin 30^\circ}$$= \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

$\frac{\sin 70^\circ + \cos 40^\circ}{\cos 70^\circ + \sin 40^\circ} = $

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