If A = sin theta |sin theta|,B = cos theta |c | vedclass

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(C) Given,$A = \sin 	heta |\sin 	heta|$,$B = \cos 	heta |\cos 	heta|$ and $\frac{99 \pi}{2} \leq 	heta \leq \frac{100 \pi}{2}$.Since $\frac{99 \p

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$B - A = 1$

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(C) Given,$A = \sin 	heta |\sin 	heta|$,$B = \cos 	heta |\cos 	heta|$ and $\frac{99 \pi}{2} \leq 	heta \leq \frac{100 \pi}{2}$.Since $\frac{99 \pi}{2} = 49 \pi + \frac{\pi}{2}$ and $\frac{100 \pi}{2} = 50 \pi$,the angle $	heta$ lies in the $4^{	ext{th}}$ quadrant.In the $4^{	ext{th}}$ quadrant,$\sin 	heta  0$.Therefore,$|\sin 	heta| = -\sin 	heta$ and $|\cos 	heta| = \cos 	heta$.Substituting these values,we get $A = \sin 	heta (-\sin 	heta) = -\sin^2 	heta$ and $B = \cos 	heta (\cos 	heta) = \cos^2 	heta$.Thus,$B - A = \cos^2 	heta - (-\sin^2 	heta) = \cos^2 	heta + \sin^2 	heta = 1$.

If $A = \sin \theta |\sin \theta|$,$B = \cos \theta |\cos \theta|$ and $\frac{99 \pi}{2} \leq \theta \leq \frac{100 \pi}{2}$,then

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