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Question : 46 of 160
Marks:
+1,
-0
Solution:
[x‌cos‌θ−y] [(cos‌θ+tan‌α)x−(1−cos‌θ‌tan‌α)y]=0 ⇒cos‌θ(cos‌θ+tan‌α)‌x2−(cos‌θ−cos2θ‌tan‌α)xy −(cos‌θ+tan‌α)yx+(1−cos‌θ‌tan‌α)y2=0 ‌⇒cos‌θ(cos‌θ+tan‌α)‌x2+xy(−cos‌θ+cos2θ.‌‌‌tan‌α−cos‌θ−tan‌α)+(1−cos‌θ‌tan‌α)y2=0 ‌⇒cos‌θ(cos‌θ+tan‌α)‌x2+(−2‌cos‌θ‌‌−sin‌2θ‌tan‌α)‌x‌y+(1−cos‌θ‌tan‌α)y2=0‌⇒cos‌θ(cos‌θ+tan‌α)‌x2−(2‌cos‌θ+sin‌2θ‌tan‌α)xy+(1−cos‌θ‌tan‌α)y2=0 We know,
tan‌θ=|‌| ‌ ‌=‌| tan‌α‌√cos4θ+2cos2θ+1 |
| 1+cos2θ |
=tan‌α‌⇒tan‌θ=tan‌α‌⇒θ=α
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