Sol. tan
θ = 1
But we know that, tanθ = y/x = 1/1
∴ y = 1 and x = 1
r = √(x2+y2
)
∴ r =√((1)2+(1)2
)
∴ r =√(1+1)
∴ r =√2
Now,
sinθ = y/r = 1/√2
cosθ = x/r = 1/√2
sec θ = r/x = √2
cosec θ = r/y = √2
∴
(sinθ + cos θ) ÷ (secθ + cosecθ ) = (1/√2 + 1√2) ÷ (√2 + √2)
∴ (sinθ + cos θ) ÷ (secθ + cosecθ ) =(1+1)/√2 ÷ 2√2
∴ (sinθ + cos θ) ÷ (secθ + cosecθ ) = 2/√2 / 2√2
∴ (sinθ + cos θ) ÷ (secθ + cosecθ ) = 2/4
∴ (sinθ + cos θ) ÷ (secθ + cosecθ ) = 1/2