frac{cos 50^{circ}}{sin 40^{circ}} + frac{sin | vedclass

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Question

(C) We use the trigonometric identity for complementary angles: $\cos 	heta = \sin(90^{\circ} - 	heta)$ and $\sin 	heta = \cos(90^{\circ} - 	heta)

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We use the trigonometric identity for complementary angles: $\cos 	heta = \sin(90^{\circ} - 	heta)$ and $\sin 	heta = \cos(90^{\circ} - 	heta)$.For the first term: $\cos 50^{\circ} = \sin(90^{\circ} - 50^{\circ}) = \sin 40^{\circ}$.Thus,$\frac{\cos 50^{\circ}}{\sin 40^{\circ}} = \frac{\sin 40^{\circ}}{\sin 40^{\circ}} = 1$.For the second term: $\cos 75^{\circ} = \sin(90^{\circ} - 75^{\circ}) = \sin 15^{\circ}$.Thus,$\frac{\sin 15^{\circ}}{\cos 75^{\circ}} = \frac{\sin 15^{\circ}}{\sin 15^{\circ}} = 1$.Adding both results: $1 + 1 = 2$.

$\frac{\cos 50^{\circ}}{\sin 40^{\circ}} + \frac{\sin 15^{\circ}}{\cos 75^{\circ}} = \ldots \ldots \ldots \ldots$

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