The value of cos (270^circ + theta ),cos (90^ | vedclass

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(D) We use the trigonometric reduction formulas:$1$. $\cos (270^\circ + 	heta) = \sin 	heta$$2$. $\cos (90^\circ - 	heta) = \sin 	heta$$3$. $\sin

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We use the trigonometric reduction formulas:$1$. $\cos (270^\circ + 	heta) = \sin 	heta$$2$. $\cos (90^\circ - 	heta) = \sin 	heta$$3$. $\sin (270^\circ - 	heta) = -\cos 	heta$Substituting these into the expression:$\cos (270^\circ + 	heta) \cos (90^\circ - 	heta) - \sin (270^\circ - 	heta) \cos 	heta$$= (\sin 	heta)(\sin 	heta) - (-\cos 	heta)(\cos 	heta)$$= \sin^2 	heta + \cos^2 	heta$Since $\sin^2 	heta + \cos^2 	heta = 1$,the final value is $1$.

The value of $\cos (270^\circ + \theta )\,\cos (90^\circ - \theta ) - \sin (270^\circ - \theta )\,\cos \theta $ is

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