Q64 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
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
- A$\frac{2}{3}, \frac{3}{2}$
- B$\frac{1}{3}, 3$
- Correct answerC$\frac{1}{3}, 3$
- D$\frac{3}{4}, \frac{4}{3}$
Solution
$\begin{aligned}
& \text { } \cos ^2 84^{\circ}+\sin ^2 126^{\circ}-\sin 84^{\circ} \cos 126^{\circ}=K \\
& \Rightarrow \frac{1+\cos 168^{\circ}}{2}+\sin ^2(90+36)^{\circ}-\frac{1}{2}\left[\sin 210^{\circ}-\sin 42^{\circ}\right]=K \\
& \Rightarrow \frac{1+\cos 168^{\circ}}{2}+\cos ^2 36^{\circ}-\frac{1}{2}\left[-\sin 30^{\circ}-\sin 42^{\circ}\right]=K \\
& \Rightarrow \frac{1+\cos 168^{\circ}}{2}+\cos ^2 36^{\circ}+\frac{1}{2} \sin 30^{\circ}+\frac{1}{2} \sin 42^{\circ}=K \\
& \Rightarrow \frac{1+\cos 168^{\circ}+\sin 42^{\circ}}{2}+\cos ^2 36^{\circ}+\frac{1}{2} \times \frac{1}{2}=K \\
& \Rightarrow \frac{1-\sin 78^{\circ}+\sin 42^{\circ}}{2}+\left(\frac{\sqrt{5}+1}{4}\right)^2+\frac{1}{4}=K \\
& \Rightarrow \frac{1}{2}-\frac{1}{2}\left(2 \cos 60^{\circ} \cdot \sin 18^{\circ}\right)+\left(\frac{\sqrt{5}+1}{4}\right)^2+\frac{1}{4}=K
\end{aligned}$
$\begin{aligned}
& \Rightarrow \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{5}-1}{4}+\frac{6+2 \sqrt{5}}{16}+\frac{1}{4}=K \\
& \Rightarrow \frac{1}{2}-\frac{\sqrt{5}}{8}+\frac{1}{8}+\frac{3}{8}+\frac{\sqrt{5}}{8}+\frac{1}{4}=K \Rightarrow K=\frac{5}{4}
\end{aligned}$
Now, $\cot A+\tan A=2 K$ $\Rightarrow \frac{1}{\tan }+\tan =\frac{5}{2}$ $\Rightarrow 2 \tan ^2 A-5 \tan A+2=0 \therefore \tan A=\frac{1}{2}, 2$.
Now, $\cot A+\tan A=2 K$ $\Rightarrow \frac{1}{\tan }+\tan =\frac{5}{2}$ $\Rightarrow 2 \tan ^2 A-5 \tan A+2=0 \therefore \tan A=\frac{1}{2}, 2$.
Hence, the correct answer is option: 'C'.