tan 2 θ + 1/tan ​2 θ=2 then value of θ=?

tan(45°) = 1   so   tan(45°) + 1/tan(45°) = 1 + 1 → 2

θ = 45°   (or θ = pi/4  radians)

.

tan^2(x) +1/tan^2(x)=2

use trig identities to solve.since

tan^2(x) =sec^2(x) -1

then write

sec^2(x)-1 +1/(sec^2(x)-1)=2

so     a + 1/a = 2     where a = sec^2(x)-1

solve quadratic in a

a^2-2a+1=0   gives (a-1)^2=0,so a=1  giving sec^2(x)=2

sec(x)=plus or minus root2

cos(x) = 1/(plus or minus root 2)

x= 45 deg, 315 deg, etc

tan²θ + 1/tan²θ=2 then value of θ=?

$\begin{array}{rcll} \tan^2(\phi) + \frac{ 1 } { \tan^2(\phi) } &=& 2 \qquad & |\qquad \cdot \tan^2(\phi)\\ \tan^4(\phi) + 1 &=& 2 \cdot \tan^2(\phi) \qquad & |\qquad -2\cdot \tan^2(\phi)\\ \tan^4(\phi) -2\cdot \tan^2(\phi) + 1 &=& 0 \qquad & |\qquad \text{binomial}\\ ( \tan^2(\phi) - 1 ) ^2 &=& 0 \qquad & |\qquad \sqrt{} \\ \tan^2(\phi) - 1 &=& 0 \qquad & |\qquad +1 \\ \tan^2(\phi) &=& 1 \qquad & |\qquad \pm\sqrt{} \\ \tan(\phi) &=& \pm\sqrt{1} \\ \tan(\phi) &=& \pm 1 \qquad & |\qquad \arctan{()} \\ \phi &=& \arctan{(\pm 1)} \pm k\cdot \pi \qquad & |\qquad k = 0,1,2,\cdots \\ \phi &=& \pm\arctan{(1)} \pm k\cdot \pi \qquad & |\qquad \arctan{(1)} = \frac{\pi}{4} \\ \phi &=& \pm\frac{\pi}{4} \pm k\cdot \pi \\ \phi_1 &=& \frac{\pi}{4} \pm k\cdot \pi \\ \phi_2 &=& -\frac{\pi}{4} \pm k\cdot \pi \\\\ \phi &=& \cdots -\frac34\pi, -\frac{\pi}{4}, \frac{\pi}{4}, \frac34\pi, \cdots \end{array}$