Q60 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics
$\begin{aligned}
\tan ^2 \frac{\pi}{16}+\tan ^2 \frac{2 \pi}{16}+\tan ^2 & \frac{3 \pi}{16}+\tan ^2 \frac{4 \pi}{16} \\
& +\tan ^2 \frac{5 \pi}{16}+\tan ^2 \frac{6 \pi}{16}+\tan ^2 \frac{7 \pi}{16}=
\end{aligned}$
- Correct answerA35
- B41
- C37
- D33
Solution
Let,
$\begin{aligned}
& K=\tan ^2 \frac{\pi}{16}+\tan ^2 \frac{2 \pi}{16}+\tan ^2 \frac{3 \pi}{16}+\ldots .+\tan ^2 \frac{7 \pi}{16} \\
& =\left\{\tan ^2 \frac{\pi}{16}+\tan ^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right)\right\}+\left\{\tan ^2\left(\frac{\pi}{8}\right)+\tan ^2\left(\frac{\pi}{2}-\frac{\pi}{8}\right)\right\} \\
& +\left\{\tan ^2 \frac{3 \pi}{2}+\tan ^2\left(\frac{\pi}{2}-\frac{3 \pi}{16}\right)\right\}+\tan ^2 \frac{\pi}{4}
\end{aligned}$
$\begin{aligned}
=\left(\tan ^2 \frac{\pi}{16}+\cot ^2 \frac{\pi}{16}\right)+\left(\tan ^2\right. & \left.\frac{\pi}{8}+\cot ^2 \frac{\pi}{8}\right) \\
& +\left(\tan ^2 \frac{3 \pi}{16}+\cot ^2 \frac{3 \pi}{16}\right)+1
\end{aligned}$
Now, $\left(\tan ^2 \frac{\pi}{16}+\cot ^2 \frac{\pi}{16}\right)=\left(\tan \frac{\pi}{16}+\cot \frac{\pi}{16}\right)^2-2$ $=\left(\frac{\sin ^2 \frac{\pi}{16}+\cos ^2 \frac{\pi}{16}}{\sin \frac{\pi}{16} \cdot \cos \frac{\pi}{16}}\right)^2-2=\frac{4}{\sin ^2 \frac{\pi}{8}}-2=\frac{8}{2 \sin ^2 \frac{\pi}{8}}-2$ $=\frac{8}{1-\cos \frac{\pi}{4}}-2=\frac{8 \sqrt{2}}{\sqrt{2}-1}-2=8 \sqrt{2}(\sqrt{2}+1)-2$ ...(i) Similarly, $\tan ^2 \frac{\pi}{8}+\cot ^2 \frac{\pi}{8}=\frac{8}{1-\cos \frac{\pi}{2}}-2=6$ ...(ii) and $\tan ^2 \frac{3 \pi}{16}+\cot ^2 \frac{3 \pi}{16}=\frac{8}{1-\cos \frac{3 \pi}{4}}-2=8 \sqrt{2}(\sqrt{2}-1)-2$ ...(iii) So, by equation (i), (ii) and (iii), we get $\begin{aligned} & K=8 \sqrt{2}(\sqrt{2}+1)-2+6+8 \sqrt{2}(\sqrt{2}-1)-2+1 \\ & =16+6+16-3=35 \end{aligned}$
Now, $\left(\tan ^2 \frac{\pi}{16}+\cot ^2 \frac{\pi}{16}\right)=\left(\tan \frac{\pi}{16}+\cot \frac{\pi}{16}\right)^2-2$ $=\left(\frac{\sin ^2 \frac{\pi}{16}+\cos ^2 \frac{\pi}{16}}{\sin \frac{\pi}{16} \cdot \cos \frac{\pi}{16}}\right)^2-2=\frac{4}{\sin ^2 \frac{\pi}{8}}-2=\frac{8}{2 \sin ^2 \frac{\pi}{8}}-2$ $=\frac{8}{1-\cos \frac{\pi}{4}}-2=\frac{8 \sqrt{2}}{\sqrt{2}-1}-2=8 \sqrt{2}(\sqrt{2}+1)-2$ ...(i) Similarly, $\tan ^2 \frac{\pi}{8}+\cot ^2 \frac{\pi}{8}=\frac{8}{1-\cos \frac{\pi}{2}}-2=6$ ...(ii) and $\tan ^2 \frac{3 \pi}{16}+\cot ^2 \frac{3 \pi}{16}=\frac{8}{1-\cos \frac{3 \pi}{4}}-2=8 \sqrt{2}(\sqrt{2}-1)-2$ ...(iii) So, by equation (i), (ii) and (iii), we get $\begin{aligned} & K=8 \sqrt{2}(\sqrt{2}+1)-2+6+8 \sqrt{2}(\sqrt{2}-1)-2+1 \\ & =16+6+16-3=35 \end{aligned}$
Hence, the correct answer is option: 'A'.