Mathematics - Quadratic Equation Question with Solution | TestHub

Given, ${32}^{{\tan }^{2}x}+{32}^{{\sec }^{2}x}=81, 0\le x\le \frac{\pi }{4}$

${32}^{{\tan }^{2}x}+{32}^{1+{\tan }^{2}x}=81$

${32}^{{\tan }^{2}x}+32.{32}^{{\tan }^{2}x}=81$

$33\times {32}^{{\tan }^{2}x}=81$

${32}^{{\tan }^{2}x}=\frac{27}{11}$

${\tan }^{2}x={\ln }_{32}\left(\frac{27}{11}\right)$
$\tan x=\sqrt{{\ln }_{32}\left(\frac{27}{11}\right)}\in \left(0,1\right)$
One solution in $\left[0,\frac{\pi }{4}\right]$