Question

The value of ${\sin }^2\frac{3\pi }{4}+{\sec }^2\frac{5\pi }{3}-{\tan }^2\frac{2\pi }{3}$ is

Answer 1

${\sin }^2\left(3\pi /4\right)+{\sec }^2\left(5\pi /3\right)-{\tan }^2\left(2\pi /3\right)$

$={\sin }^2(\pi -\pi /4)+{\sec }^2(2\pi -\frac{\pi }{3})-{\tan }^2(\pi -\frac{\pi }{3})$ $\left[\begin{array}{c}\sin (\pi -x)=\sin x\\ \sec (2\pi -x)=\sec x\\ \tan (\pi -x)=-\tan x\end{array}\right]$

$=\left(\sin \right(\pi /4){)}^2+(\sec \left(\pi /3\right){)}^2-(-\tan (\pi /3){)}^2$

$={\left(\frac{1}{\sqrt{2}}\right)}^2+(2{)}^2-(-\sqrt{3}{)}^2$

$=\frac{1}{2}+4-3$

$=\frac{1+8-6}{2}=\frac{3}{2}$

$\therefore {\sin }^2\left(3\pi /4\right)+{\sec }^2\left(5\pi /3\right)-{\tan }^2\left(2\pi /3\right)=3/2$.