If sec θ + tan θ = 2 + √5 and θ is an acute angle, then the value of…

We know ,

sec²θ - tan²θ = 1

So , secθ - tanθ =  \(\frac{1 }{secθ + tanθ }\)

ATQ, 

secθ + tanθ = 2 + √5    ------(1)

So, secθ - tanθ =  \(\frac{1 }{2 + √5 }\)

= \(\frac{1 }{2 + √5 }\) × \(\frac{2 - √5 }{2 - √5 }\)

= √5 - 2       --------(2)

Adding 1 and 2

2 secθ = 2√5

cosθ = \(\frac{1 }{ √5 }\) 

We know ,

sin²θ + cos²θ = 1

sin²θ + \(\frac{1 }{ 5 }\)  = 1

sinθ = \(\frac{2 }{ √5 }\)

= \(\frac{2 √5}{ 5 }\)