cos⁡tan−1⁡sin⁡cot−1⁡x=

Let cot−1⁡x=θ⇒cot⁡θ=x∴sin⁡cot−1⁡x=sin⁡θ=1cosec⁡θ=11+cot2⁡θ=11+x2Thus, cos⁡tan−1⁡sin⁡cot−1⁡x=cos⁡tan−1⁡11+x2=cos⁡φ Put tan−1⁡11+x2=ϕ⇒tan⁡ϕ=11+x2=1sec⁡ϕ=11+tan2⁡ϕ=11+11+x2=1+x22+x2