Question

The value of $anti{\log }_5\left[\frac{{\tan }^2\left(\frac{\pi }{5}\right)+{\tan }^2\left(\frac{2\pi }{5}\right)+20}{{\cot }^2\left(\frac{\pi }{5}\right)+{\cot }^2\left(\frac{2\pi }{5}\right)+28}\right]$ is equal to

• A. odd number
• B. even number
• C. prime number
• D. composite number

Answer 1

Answer: (A), (C)

The correct options are
A odd number
C prime number

$\frac{{tan}^2\left(\frac{\Pi }{5}\right)+{tan}^2\left(\frac{2\Pi }{5}\right)+20}{{cot}^2\left(\frac{\Pi }{5}\right)+{cot}^2\left(\frac{2\Pi }{5}\right)+28}=\frac{{\left(\sqrt{5-2\sqrt{5}}\right)}^2+(\sqrt{\frac{5}{5-2\sqrt{5}}}{)}^2+20}{{\left(\frac{1}{\sqrt{5-2\sqrt{5}}}\right)}^2+{(\sqrt{\frac{5-2\sqrt{5}}{5})}}^2+28}=\frac{5-2\sqrt{5}+\frac{5}{5-2\sqrt{5}}+20}{\frac{1}{5-2\sqrt{5}}+\frac{5-2\sqrt{5}}{5}+28}=\frac{5\left[\right(5-{2\sqrt{5})}^2+5+20(5-2\sqrt{5})]}{5+(5-{2\sqrt{5})}^2+140(5-2\sqrt{5})}=\frac{5(45-20\sqrt{5}+5+100-40\sqrt{5})}{5+45-20\sqrt{5}+700-200\sqrt{5}}=\frac{750-300\sqrt{5}}{750-300\sqrt{5}}antilo{g}_5\left[\frac{{tan}^2(\frac{\Pi }{5}+{tan}^2(\frac{2\Pi }{5})+20}{{cot}^2\left(\frac{\Pi }{5}\right)+{cot}^2\left(\frac{2\Pi }{5}\right)+28}\right]=5^1=5$