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Question : 116 of 200
Marks:
+1,
-0
Solution:
Given that,
‌tan‌θ+sin‌θ=x‌tan‌θ−sin‌θ=y‌tan‌θ=‌,sin‌θ=‌∴‌‌(‌)(‌)=tan‌θ‌s‌i‌n‌θ⇒‌‌x2−y2=4(‌)=4(‌)⇒‌‌x2−y2=4( secθ−cos‌θ)⇒‌‌(x2−y2)2=16( secθ2+cos2θ−2)⇒‌‌(x2−y2)2=16(1+tan2θ+1−sin‌2θ−2)⇒‌‌(x2−y2)2=16(tan2θ−sin‌2θ)⇒‌‌(x2−y2)2=16xy
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