if sin2X=k then prove that (tan^3X)/(1+tan^2X)+(Cot^3X)/(1+cot^2X)=(2-k^2)/k
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This implies that
x2+2ax=4x−4a−13
or
x2+2ax−4x+4a+13=0
or
x2+(2a−4)x+(4a+13)=0
Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.
Hence we get that
(2a−4)2=4⋅1⋅(4a+13)
or
4a2−16a+16=16a+52
or
4a2−32a−36=0
or
a2−8a−9=0
or
(a−9)(a+1)=0
So the values of a are −1 and 9.
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