costan−1sincot−1x=
x2+2x2+3
x2+2x2+1
x2+1x2+2
none of these
Let cot−1x=θ⇒cotθ=x
∴sincot−1x=sinθ=1cosecθ=11+cot2θ=11+x2
Thus, costan−1sincot−1x
=costan−111+x2=cosφ
Put tan−111+x2=ϕ⇒tanϕ=11+x2
=1secϕ=11+tan2ϕ=11+11+x2=1+x22+x2