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Question : 75 of 160
Marks:
+1,
-0
Solution:
‌(√3−1)sin‌θ+(√3+1)‌cos‌θ=2‌‌sin‌θ+‌‌cos‌θ=1 . . . (i)
Comparing with
asin‌θ+b‌cos‌θ=1.
ie,
a‌=‌,b=‌√a2+b2‌=√‌+‌‌=‌√3+1−2√3+3+1+2√3‌=‌√8=‌⋅2√2=√2Dividing on both sides by
√2 in Eq. (i), we get
(‌)sin‌θ+(‌)‌cos‌θ=‌Let
sin‌α=‌‌ Then, ‌cos‌α‌=√1−(‌)2‌=√1−‌‌=√‌=√‌‌=√‌=√‌‌=(‌) ‌‌ So, ‌‌‌sin‌α⋅sin‌θ+cos‌α⋅cos‌θ=‌‌cos(θ−α)=‌=cos‌‌θ−α=2nπ±‌,θ=2nπ±‌+α . . . (ii)
‌∵‌‌cos‌15=‌‌‌ ie, ‌‌‌cos‌α=cos‌15∘=cos‌‌⇒‌‌α=‌From Eq. (ii),
[θ=2nπ±‌+‌],n∈Z
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