If cos⁡α=12x+1x and cos⁡β=12y+1y,(xy>0) ;x,y,α,β,∈R then

cos⁡α=12x+1x and cos⁡β=12y+1ysince xy > 0, we havex+1x≥2 or ≤−2 and y+1y≥2 or ≤−2⇒cos⁡α=1,cos⁡β=1----1or  cos⁡α=−1,cos⁡β=−1----2cos⁡αcos⁡β=1⇒α+β is an even multiple of π(cos⁡α+cos⁡β)2=4⇒sin⁡(α+β+γ)=sin⁡(2nπ+γ)=sin⁡γAlso ,sin⁡α=sin⁡β=0.