cosθ−sinθ=15
Squaring both sides, we get
1−2sinθcosθ=125⇒2sinθcosθ=2425
Now,
(sinθ+cosθ)2=1+2sinθcosθ=1+2425=4925⇒sinθ+cosθ=±75
As θ∈(0,π2), so
sinθ+cosθ=75∴5(sinθ+cosθ)=7
Alternate method:
cosθ−sinθ=15
Assuming sinθ+cosθ=x
Squaring and adding both the equations, we get
1−2sinθcosθ+1+2sinθcosθ=x2+125⇒x2+125=2⇒x=±75
As θ∈(0,π2), so
sinθ+cosθ=75∴5(sinθ+cosθ)=7