EXAMPLE 3 140
Question
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EXAMPLE 3 140
JEE/Engineering Exams
Maths
Solution
86
Rating
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y+11-y=1+sinΘ1-sin2 =sin2Θ/2+cos2Θ/2+2sinΘ/2cosΘ/2sin2Θ/2+cos2Θ/2-2sinΘ/2cosΘ/2 =(cosΘ/2+sinΘ/2)2cosΘ2-sinΘ22 =cosΘ/2+sinΘ/2cosΘ/2-sinΘ/2 =1+tanΘ/21-tanΘ/2 Use compendo & dividends rule =y+1+1-yy+1-1+y=1+tanΘ/2+1-tanΘ/21+tanΘ/2-1+tanΘ/2 22y=22tanΘ/2 y=tanΘ2

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