If θ is an acute angle such that sec²θ = 3, then the value of is

Given,

Sec²θ = 3

So,

Sec θ = √3 = h/b = k

Therefore,

h = √3k, b = k

In ∆ABC,

h² = p² + b²

(√3k)² = p² + (k)²

3k² = p² + k²

3k² – k² = p²

2k² = p²

√2 K = p

We know,

tan θ = p/b = √2k/k = √2

cosec θ = h/p = √3k/√2k = √3/√2

Put these value in,