If r sinθ = 1, r cosθ = sqrt 3 , then value of (r2 tanθ) is?

r sinθ = 1 ----- (i)

r cosθ = \(\sqrt {3 }\)

squaring and adding

r²sin²θ + r²cos²θ = 4

⇒ r²(sin²θ + cos²θ) = 4

⇒ r² = 4 

⇒ r = 2

divide (i) by (ii)

tanθ = \(\frac{1}{\sqrt {3 }}\)

r²tanθ = \(\frac{4}{\sqrt {3 }}\)