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If `cos (x-y),, cosx and cos(x+y)` are in H.P., are in H.P., then `cosx*sec(y/2)`=A. `-sqrt(3)`B. `-sqrt(2)`C. `sqrt(2)`D. `sqrt(3)` |
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Answer» Correct Answer - B::C We have, `" " (2)/(cosx ) = (1)/(cos(x-y)) + (1)/(cos(x+y)) = ( 2cosx*cosy)/( cos^(2)x - sin^(2)y)` `rArr cos^(2) x - sin^(2)y = cos^(2) x * cos y` `rArr cos^(2)x ( 1-cosy) = sin^(2)y` `rArr cos^(2) x* 2 sin^(2)""(y)/(2) = 4 sin^(2) ""(y)/(2)cos^(2)""(y)/(2)` `rArr cos ^(2)x= 2 cos^(2)""(y)/(2)` `rArr cos^(2)x sec^(2) ""(y)/(2)= 2` `rArr cosx * sec""(y)/(2) = pm sqrt2` |
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