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,