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question mark

Pls find the attachment
Wat would be its value
Pls reply fast

Nidhi Bhat , 9 Years ago
Grade 11
anser 1 Answers
Vikas TU

Last Activity: 9 Years ago

Taking first,
(cos^3 A + sin^3 A)/(cosA + sinA) = > (cosA + sinA) (cos^2 A + sin^2 A – cosAsinA)/(cosA + sinA)
                                                  => (1 – cosAsinA)
Taking now,
cos^3 A . sin^3 A/(cosA.sinA) = > (cosA.sinA)^2
 
Taking together,
(cosA.sinA)^2 +  (1 – cosAsinA)
let cosAsinA = t
or
sin2A = 2t
    -1
            or
          -1/2
but t^2  –  t  + 1 > 0 
.i.e D
t cannot be imaginary.
Hence no solution.
 
 
NotE: Plz check that second term in yr. attached image i think there is misconception in denominator.
thnx!

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