If $cosec^2θ + cot^2 θ = \frac{5}{3}$, then what is the value of cot2θ ?
Answer & explanation
Correct answer: option 2
cosec²θ + cot²θ = \(\frac{5}{3}\) ----(1)
cosec²θ - cot²θ = 1 ----(2)
On subtracting equation 2 from equation 1.
2 cot²θ = \(\frac{5}{3}\) - 1
cot²θ = \(\frac{1}{3}\)
cotθ = \(\frac{1}{√3}\)
{ we know, cot60º = \(\frac{1}{√3}\) }
So,
θ = 60º
Now,
cot2θ
= cot ( 2 × 60º)
= cot 120º
= cot ( 180º - 60º )
{ cot is negative in 2nd quadrant }
= - cot120º
= - \(\frac{1}{√3}\)