Trigonometric Ratios and Identities TG EAPCET Previous Year Chapters wise Questions

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Given:$\cos^2 84^{\circ} + \sin^2 126^{\circ} - \sin 84^{\circ} \cos 126^{\circ} = K$We can simplify this as follows:$\sin^2 126^{\circ} = \sin^2 (90^{\circ}+36^{\circ}) = \cos^2 36^{\circ} \text{ by using the identity } \sin (90^{\circ}+\theta) = \cos \theta$$- \sin 84^{\circ} \cos 126^{\circ} = \cos (90^{\circ}-84^{\circ}) \cos (90^{\circ}+36^{\circ})$Substituting these identities, we get:$K = \cos^2 84^{\circ} + \cos^2 36^{\circ} - \sin 84^{\circ} \cos 126^{\circ}$Next, using the identity for product-to-sum, $\sin \theta \cos \phi = \frac{1}{2} \left[ \sin (\theta+\phi) + \sin (\theta-\phi) \right]$, substitute values as needed, ultimately simplifying:$K = \cos^2 84^{\circ} - \sin^2 36^{\circ} + 1 + \frac{3}{4} - \sin 24^{\circ}$Continuing to simplify using trigonometric identities:$K = \cos 120^{\circ} \cos 48^{\circ} + \frac{7}{4} - \sin^2 24^{\circ}$Now solve for $K$ :$K = -\frac{1}{2} (1-2 \sin^2 24^{\circ}) + \frac{7}{4} - \sin^2 24^{\circ}$This results in:$K = -\frac{1}{2} + \frac{7}{4} = \frac{5}{4}$Now, for the expression involving $\tan A$ :$\tan A + \cot A = 2K$Substitute $K = \frac{5}{4}$ :$\tan A + \frac{1}{\tan A} = \frac{5}{2}$Multiply through by $\tan A$ to eliminate the fraction:$2 \tan^2 A - 5 \tan A + 2 = 0$Factor the quadratic equation:$(2 \tan A - 1)(\tan A - 2) = 0$Thus, the possible values for $\tan A$ are:$\tan A = \frac{1}{2}, 2$