if tan(πcosθ)=cot(πsinθ)
then form eqn whose roots are cosθ and sinθ
1tan { Î cos x } = tan { ( 2n + 1 ) Î / 2 - Î sin x }
so , Î cos x = nÎ + { nÎ + Î / 2 - Î sin x }
or , cos x = 2 n + 1 / 2 - sin x
or , cos x + sin x = { 4 n + 1 } / 2 ,
but , sin x + cos x ≤ √2
so , n = 0 ,
That gives , cos x + sin x = 1 / 2
or , √2 sin ( x + Î / 4 ) = 1 / 2
or , x + Î / 4 = sin - 1 { 1 / 2√2 }
or , x = Î / 4 - sin - 1 { 1 / 2√2 }
Now we can find out cos x and sin x , and the corresponding equation .
1@ gallardo u have given just a specific solution
this is a famous sl loney problem
just write
tan (πcosθ)=tan(π2-πsinθ)
general solution is
πcosθ=nπ+(π2-πsinθ)
cancelling π
cosθ=n+12-sinθ ..................1
√2(cos(Ï€/4-θ))=n+12
(cos(θ-Ï€4))=1√2(n+12)
θ=Ï€4+2nπ±cos-1(1√2(n+12))
now use this to get your answer
