If tan θ + cot θ = 6, then find the value of $$\text{tan}^2\text{θ}+\text{cot}^2\text{θ}$$​

Correct option is B

Given:
tan θ + cot θ = 6 
Formula Used:
​$$\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$$​​
Solution:
Let x = tan θ => cot θ = $$\frac 1x$$​
tan θ + cot θ = 6 
​$$x + \frac{1}{x}$$​ = 6
Squaring both sides
​$$\left(x + \frac{1}{x}\right)^2 = 6^2 = 36 \\x^2 + 2 + \frac{1}{x^2} = 36 \\x^2 + \frac{1}{x^2} = 36 - 2 = 34$$​​