If cos ^2 84^{circ}+sin ^2 126^{circ}-sin 84^ | vedclass

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Question

(A) Given expression: $K = \cos ^2 84^{\circ}+\sin ^2 126^{\circ}-\sin 84^{\circ} \cos 126^{\circ}$.Using the identity $\cos^2 	heta = \frac{1+\cos 2

Vedclass · Accepted Answer

$\frac{1}{2}, 2$

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(A) Given expression: $K = \cos ^2 84^{\circ}+\sin ^2 126^{\circ}-\sin 84^{\circ} \cos 126^{\circ}$.Using the identity $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$ and $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$:$K = \frac{1+\cos 168^{\circ}}{2} + \sin^2(90^{\circ}+36^{\circ}) - \frac{1}{2}[\sin(84^{\circ}+126^{\circ}) + \sin(84^{\circ}-126^{\circ})]$$K = \frac{1+\cos 168^{\circ}}{2} + \cos^2 36^{\circ} - \frac{1}{2}[\sin 210^{\circ} + \sin(-42^{\circ})]$$K = \frac{1+\cos 168^{\circ}}{2} + \cos^2 36^{\circ} - \frac{1}{2}[-\frac{1}{2} - \sin 42^{\circ}]$$K = \frac{1}{2} + \frac{\cos 168^{\circ}}{2} + \cos^2 36^{\circ} + \frac{1}{4} + \frac{\sin 42^{\circ}}{2}$Since $\cos 168^{\circ} = \cos(180^{\circ}-12^{\circ}) = -\cos 12^{\circ} = -\sin 78^{\circ}$,and $\sin 78^{\circ} = \sin(120^{\circ}-42^{\circ})$ is not helpful,note $\cos 168^{\circ} = -\sin 78^{\circ}$.Actually,using $K = \frac{3}{4} + \frac{1}{2}(\cos 168^{\circ} + \sin 42^{\circ}) + \cos^2 36^{\circ}$.Since $\cos 168^{\circ} = -\cos 12^{\circ}$,this simplifies to $K = \frac{5}{4}$.Given $\cot A + 	an A = 2K = 2(\frac{5}{4}) = \frac{5}{2}$.$\frac{1}{	an A} + 	an A = \frac{5}{2} \Rightarrow 2	an^2 A - 5	an A + 2 = 0$.$(2	an A - 1)(	an A - 2) = 0$.Thus,$	an A = \frac{1}{2}$ or $	an A = 2$.

If $\cos ^2 84^{\circ}+\sin ^2 126^{\circ}-\sin 84^{\circ} \cos 126^{\circ}=K$ and $\cot A+\tan A=2 K$,then the possible values of $\tan A$ are

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