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Let `alpha, beta` be such that `pi lt alph-betalt3piif sin alpha+sinbeta=-21/65and cos alpha+cos beta =-27/65,` then the value of `cos""(alpha-beta)/(2),` isA. `-(6)/(65)`B. `(3)/(sqrt130)`C. `6/65`D. `-(3)/(sqrt130)` |
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Answer» Correct Answer - D We have, `sin alpha+sinbeta=-21/65and cos alpha+cosbeta=-27/65` `implies2sin""(alpha+beta)/(2)cos""(alpha-beta)/(2)=-21/65` `and 2cos""(alpha+beta)/(2)cos""(alpha-beta)/(2)=-27/65` `implies4(sin^(2)""(alpha+beta)/(2)+cos^(2)""(alpha+beta)/(2))cos^(2)""(alpha-beta)/(2)=(-(21)/(65))^(2)+(-(27)/(65))^(2)` `implies4cos^(2)((alpha-beta)/(2))=(1170)/(65^(2))` `implies4cos^(2)((alpha-beta)/(2))=(1170)/(4xx65^(2))` `implies|cos""(alpha-beta)/(2)|=(sqrt(1170))/(130)` `impliescos""(alpha-beta)/(2)=-(sqrt(1170))/(130)" "[{:(,because(pi)/(2)lt(alpha-beta)/(2)lt(3pi)/(2)),(,thereforecos""(alpha-beta)/(2)lt0):}]` `impliescos""(alpha-beta)/(2)=-sqrt((1170)/(130xx130))=-(3)/(sqrt(130))` |
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