cos⁡tan−1⁡sin⁡cot−1⁡x=

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x2+2x2+3

x2+2x2+1

x2+1x2+2

none of these

answer is C.

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Detailed Solution

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

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