The value of the summation is =?

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1 Answer
Mar 21, 2018

The Right Option is (4): -4-2sqrt3.

Explanation:

|(0,cosx,-sinx),(sinx,0,cosx),(cosx,sinx,0)|=0.

Expanding by the first raw, we have,

:. 0-cosx(0-cos^2x)-sinx(sin^2x-0)=0.

:. cos^3x-sin^3x=0, or, sin^3x=cos^3x......(star).

We must have cos^3x!=0, because," otherwise "cosx=0, and, then,

(star) rArr sin^3x=cos^3x=0," so that, "sinx=0.

rArr cos^2x+sin^2x=0+0=0, a contradiction.

So, dividing (star)" by "cos^3x!=0, "we get, "tan^3x=1.

:. tanx=1, &, x in [0,2pi], x=pi/4,or, x=pi/4+pi=5/4pi.

:. S={pi/4,5pi/4} rArr AA x in S, tanx=1.

:. sum_( x in S)tan(pi/3+x),

=tan(pi/3+pi/4)+tan(pi/3+5pi/4),

={tan(pi/3)+tan(pi/4)}/{1-tan(pi/3)tan(pi/4)}

+{tan(pi/3)+tan(5pi/4)}/{1-tan(pi/3)tan(5pi/4)},

=(sqrt3+1)/(1-sqrt3)+(sqrt3+1)/(1-sqrt3),

={2(sqrt3+1)}/(1-sqrt3)xx(1+sqrt3)/(1+sqrt3),

=2/{1-3)(1+sqrt3)^2.

rArr sum_( x in S)tan(pi/3+x)=-(1+2sqrt3+3),

showing that the right option is (4): -4-2sqrt3.

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