cos⁻¹(n/(n+1))
cos⁻¹(n²/(n²+1))
sin⁻¹(n/(n+1))
sin⁻¹(n²/(n²+1))
|A+B|=n|A−B|⇒(A²+B²+2ABcosθ)=n²(A²+B²−2ABcosθ)It is given that |A|=|B|Hence cosθ=(n²−1)/(n²+1) ⇒θ=cos⁻¹((n²−1)/(n²+1))