cos (40^{circ}-theta)-sin (50^{circ}+theta) = | vedclass

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Question

(D) આપણે જાણીએ છીએ કે $\sin(90^{\circ}-A) = \cos A$.તેથી,$\sin(50^{\circ}+	heta) = \cos(90^{\circ}-(50^{\circ}+	heta))$.$= \cos(90^{\circ}-50^{\circ

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Vedclass · Answer

(D) આપણે જાણીએ છીએ કે $\sin(90^{\circ}-A) = \cos A$.તેથી,$\sin(50^{\circ}+	heta) = \cos(90^{\circ}-(50^{\circ}+	heta))$.$= \cos(90^{\circ}-50^{\circ}-	heta) = \cos(40^{\circ}-	heta)$.આ કિંમતને મૂળ પદાવલિમાં મૂકતા:$\cos(40^{\circ}-	heta) - \sin(50^{\circ}+	heta) = \cos(40^{\circ}-	heta) - \cos(40^{\circ}-	heta) = 0$.

$\cos (40^{\circ}-\theta)-\sin (50^{\circ}+\theta) = \ldots \ldots \ldots \ldots$

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