If tan theta - cot theta = a and sin theta + | vedclass

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(D) Given that $	an 	heta - \cot 	heta = a$ $(i)$ and $\sin 	heta + \cos 	heta = b$ $(ii)$.Now,consider the expression $({b^2} - 1)^2({a^2} + 4)$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given that $	an 	heta - \cot 	heta = a$ $(i)$ and $\sin 	heta + \cos 	heta = b$ $(ii)$.Now,consider the expression $({b^2} - 1)^2({a^2} + 4)$.Since $b = \sin 	heta + \cos 	heta$,we have $b^2 = (\sin 	heta + \cos 	heta)^2 = 1 + 2 \sin 	heta \cos 	heta = 1 + \sin 2	heta$.Thus,$b^2 - 1 = \sin 2	heta$,so $(b^2 - 1)^2 = \sin^2 2	heta$.Since $a = 	an 	heta - \cot 	heta$,we have $a^2 = (	an 	heta - \cot 	heta)^2 = 	an^2 	heta + \cot^2 	heta - 2$.Thus,$a^2 + 4 = 	an^2 	heta + \cot^2 	heta + 2 = (	an 	heta + \cot 	heta)^2$.Substituting these into the expression:$(b^2 - 1)^2(a^2 + 4) = \sin^2 2	heta (	an 	heta + \cot 	heta)^2$$= (2 \sin 	heta \cos 	heta)^2 \left( \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} ight)^2$$= 4 \sin^2 	heta \cos^2 	heta \left( \frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} ight)^2$$= 4 \sin^2 	heta \cos^2 	heta \left( \frac{1}{\sin 	heta \cos 	heta} ight)^2$$= 4 \sin^2 	heta \cos^2 	heta \cdot \frac{1}{\sin^2 	heta \cos^2 	heta} = 4$.

If $\tan \theta - \cot \theta = a$ and $\sin \theta + \cos \theta = b,$ then ${({b^2} - 1)^2}({a^2} + 4)$ is equal to

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