Question

tan(π/4+1/2cos^-1a/b)+tan(π/4-1/2cos^-1a/b)=?

Answer 1

L.H.S. tan[π4+12cos−1ab]+tan[π4−12cos−1ab]tan⁡[π4+12cos−1ab]+tan⁡[π4−12cos−1ab]

Let cos−1ab=θ,cos−1ab=θ, then cosθ=abcos⁡θ=ab and 12cos−1ab=θ212cos−1ab=θ2

= tan[π4+θ2]+tan[π4−θ2]tan⁡[π4+θ2]+tan⁡[π4−θ2]

= tanπ4+tanθ21−tanπ4tanθ2+tanπ4−tanθ21+tanπ4tanθ2tan⁡π4+tan⁡θ21−tan⁡π4tan⁡θ2+tan⁡π4−tanθ21+tan⁡π4tan⁡θ2

= 1+tanθ21−tanθ2+1−tanθ21+tanθ21+tan⁡θ21−tan⁡θ2+1−tanθ21+tan⁡θ2

= (1+tanθ2)2+(1−tanθ2)21−tan2θ2(1+tan⁡θ2)2+(1−tan⁡θ2)21−tan2θ2

= 2(1+tan2θ2)1−tan2θ22(1+tan2θ2)1−tan2θ2

= 2(1cosθ)2(1cos⁡θ)

= 2ba2ba

= R.H.S.

 

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