We have, tan(5πcosθ)=cot(5πsinθ)⇒tan(5πcosθ)=tan(2π−5πsinθ)⇒5πcosθ+5πsinθ=2π⇒cosθ+sinθ=101 On multiplying both sides by 21, we get 2cosθ+2sinθ=1021⇒cosθsin4π+sinθcos4π=1021⇒sin(4π+θ)=1021(∵sin(A+B)=sinAcosB+cosAsinB)∴ only 2 solutions are there in (0,2π)