If tan θ + cot θ = 4, then the ratio of $3(tan^2\, θ + cot^2\, θ) \, to \, (2\, cosec^2\, θ\, sec^2\, θ - 4) $ will be :
Answer & explanation
Correct answer: option 4
tanθ + cotθ = 4
On squaring both side
(tanθ + cotθ)² = 4²
tan²θ + cot²θ + 2tanθ.cotθ = 16
tan²θ + cot²θ = 16- 2 { tanθ = \(\frac{1}{cotθ}\)
tan²θ + cot²θ = 14
Now,
$3(tan^2\, θ + cot^2\, θ) \, to \, (2\, cosec^2\, θ\, sec^2\, θ - 4) $
3 (tan²θ + cot²θ ) : 2 ( 1 + cot²θ ) ( 1 + tan²θ ) - 4
3 (tan²θ + cot²θ ) : 2 ( 1 + cot²θ+ tan²θ + 2cot²θ tan²θ ) - 4
3 ( 14 ) : 2 × ( 1 + 14 + 1 ) - 4
42 : 28
3 : 2