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

$4$

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(D) Given that $	an 	heta - \cot 	heta = a$ $(i)$ and $\sin 	heta + \cos 	heta = b$ $(ii)$.Now,consider the expression $({b^2} - 1)^2({a^2} + 4)$.Substitute $b = \sin 	heta + \cos 	heta$:$b^2 - 1 = (\sin 	heta + \cos 	heta)^2 - 1 = 1 + 2 \sin 	heta \cos 	heta - 1 = 2 \sin 	heta \cos 	heta = \sin 2	heta$.So,$(b^2 - 1)^2 = \sin^2 2	heta$.Substitute $a = 	an 	heta - \cot 	heta$:$a^2 + 4 = (	an 	heta - \cot 	heta)^2 + 4 = 	an^2 	heta + \cot^2 	heta - 2 + 4 = 	an^2 	heta + \cot^2 	heta + 2 = (	an 	heta + \cot 	heta)^2$.Since $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin 2	heta}$.Therefore,$(a^2 + 4) = (\frac{2}{\sin 2	heta})^2 = \frac{4}{\sin^2 2	heta}$.Multiplying the two parts:$(b^2 - 1)^2(a^2 + 4) = \sin^2 2	heta 	imes \frac{4}{\sin^2 2	heta} = 4$.Alternatively,since the expression is independent of $	heta$,let $	heta = 45^\circ$. Then $a = 	an 45^\circ - \cot 45^\circ = 1 - 1 = 0$ and $b = \sin 45^\circ + \cos 45^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.Substituting these values: $((\sqrt{2})^2 - 1)^2(0^2 + 4) = (2 - 1)^2(4) = 1^2 	imes 4 = 4$.

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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