Given, (√3−1)‌sin‌θ+(√3+1)‌cos‌θ=2 Let ‌‌√3−1=r‌sin‌α . . . (i) and √3+1=r‌cos‌α ∴r2(sin2α+cos2α)=(√3−1)2+(√3+1)2 ⇒r2=3+1−2√3+3+1+2√3 ⇒‌‌r2=8⇒r=2√2 and tan‌α=‌
√3−1
√3+1
=‌
1−1∕√3
1+1∕√3
‌‌=tan(‌
Ï€
4
−‌
Ï€
6
)=tan(‌
Ï€
12
)⇒α=‌
Ï€
12
From Eq. (i), we have r‌sin‌α‌sin‌θ+r‌cos‌α‌cos‌θ=2 ⇒r‌cos(θ−α)=2 ⇒2√2‌cos(θ−α)=2 ⇒cos(θ−α)=‌