sin^{2} 60^{circ} - tan 45^{circ} + cos^{2} 3 | vedclass

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Question

(A) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$	an 45^{\circ} = 1$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,and $\cot 90

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$	an 45^{\circ} = 1$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,and $\cot 90^{\circ} = 0$.Substituting these values into the expression:$\sin^{2} 60^{\circ} - 	an 45^{\circ} + \cos^{2} 30^{\circ} - \cot 90^{\circ} = \left(\frac{\sqrt{3}}{2}ight)^{2} - 1 + \left(\frac{\sqrt{3}}{2}ight)^{2} - 0$$= \frac{3}{4} - 1 + \frac{3}{4} - 0$$= \frac{3}{4} + \frac{3}{4} - 1$$= \frac{6}{4} - 1 = \frac{3}{2} - 1 = \frac{1}{2}$.

$\sin^{2} 60^{\circ} - \tan 45^{\circ} + \cos^{2} 30^{\circ} - \cot 90^{\circ} = \ldots$

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