If a cos theta + b sin theta = p and a sin th | vedclass

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(B) Given equations are:$a \cos 	heta + b \sin 	heta = p$ ....$(1)$$a \sin 	heta - b \cos 	heta = q$ ....$(2)$Squaring both equations:$(a \cos 	h

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$a^{2} + b^{2} = p^{2} + q^{2}$

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(B) Given equations are:$a \cos 	heta + b \sin 	heta = p$ ....$(1)$$a \sin 	heta - b \cos 	heta = q$ ....$(2)$Squaring both equations:$(a \cos 	heta + b \sin 	heta)^{2} = p^{2}$$a^{2} \cos^{2} 	heta + b^{2} \sin^{2} 	heta + 2ab \sin 	heta \cos 	heta = p^{2}$ ....$(3)$$(a \sin 	heta - b \cos 	heta)^{2} = q^{2}$$a^{2} \sin^{2} 	heta + b^{2} \cos^{2} 	heta - 2ab \sin 	heta \cos 	heta = q^{2}$ ....$(4)$Adding equation $(3)$ and $(4)$:$(a^{2} \cos^{2} 	heta + a^{2} \sin^{2} 	heta) + (b^{2} \sin^{2} 	heta + b^{2} \cos^{2} 	heta) + (2ab \sin 	heta \cos 	heta - 2ab \sin 	heta \cos 	heta) = p^{2} + q^{2}$$a^{2}(\cos^{2} 	heta + \sin^{2} 	heta) + b^{2}(\sin^{2} 	heta + \cos^{2} 	heta) = p^{2} + q^{2}$Since $\sin^{2} 	heta + \cos^{2} 	heta = 1$,we get:$a^{2} + b^{2} = p^{2} + q^{2}$

If $a \cos \theta + b \sin \theta = p$ and $a \sin \theta - b \cos \theta = q$,then the relation between $a, b, p,$ and $q$ is:

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