Prove the following : secθ+secϕ+(tanθ−tanϕ)secθ+secϕ−(tanθ−tanϕ)=tan(π4+θ2)tan(π4+ϕ2)
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Solution
L.H.S=(cosθ+cosϕ)+sin(θ−ϕ)(cosθ+cosϕ)−sin(θ−ϕ) =cosθ+ϕ2+sinθ−ϕ2cosθ+ϕ2−sinθ−ϕ2 ....(1) we have cancelled the common factor 2 cos θ−ϕ2 Now replace cos θ+ϕ2 by sin [π2−(θ−π2)] and apply sin C ± sin D etc