frac{cos ^{2} 40^{circ}+cos ^{2} 50^{circ}}{s | vedclass

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(C) We know that $\cos(90^{\circ}-	heta) = \sin 	heta$ and $\sin(90^{\circ}-	heta) = \cos 	heta$.Therefore,$\cos^{2} 50^{\circ} = \cos^{2}(90^{\ci

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$1$

Vedclass · Answer

(C) We know that $\cos(90^{\circ}-	heta) = \sin 	heta$ and $\sin(90^{\circ}-	heta) = \cos 	heta$.Therefore,$\cos^{2} 50^{\circ} = \cos^{2}(90^{\circ}-40^{\circ}) = \sin^{2} 40^{\circ}$.Similarly,$\sin^{2} 50^{\circ} = \sin^{2}(90^{\circ}-40^{\circ}) = \cos^{2} 40^{\circ}$.Substituting these values into the expression:$\frac{\cos^{2} 40^{\circ} + \sin^{2} 40^{\circ}}{\sin^{2} 40^{\circ} + \cos^{2} 40^{\circ}}$.Using the trigonometric identity $\sin^{2} 	heta + \cos^{2} 	heta = 1$,we get:$\frac{1}{1} = 1$.

$1. \cos \theta$	$a. \frac{\cos \theta}{\sin \theta}$
$2. \tan \theta$	$b. \frac{1}{\csc \theta}$
$3. \cot \theta$	$c. \frac{1}{\sec \theta}$
$4. \sin \theta$	$d. \frac{1}{\cot \theta}$
	$e. \sin \theta \cdot \cos \theta$

$\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}=\ldots \ldots \ldots \ldots$

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