Question #7318f

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Answer 1 · 2016-07-14T07:41:29

The question has not been written properly.

Given
$csc x = 3, where 90^{\circ} < x < 180^{\circ}$ $cscx=3," where " 90^@ < x < 180^@$

So $45^{\circ} < \frac{x}{2} < 90^{\circ}$ $45^@ < x/2 < 90^@$

$\Rightarrow sin x = \frac{1}{3}$ $=>sinx=1/3$

$90^{\circ} < x < 180^{\circ} \to cos x is negative$ $90^@ < x < 180^@->cosx" is negative"$

$:.cosx=-sqrt(1-sin^2x)=-sqrt(1-1/9)$
$=-(2sqrt2)/3$

$"Now "cos(x/2)=sqrt(1/2(1+cosx))$

$=sqrt(1/2(1-(2sqrt2)/3))$

$=sqrt(1/6(3-2sqrt2))$

$=sqrt(1/6((sqrt2)^2+1^2-2sqrt2*1)$

$=sqrt(1/6(sqrt2-1)^2$

$=1/sqrt6(sqrt2-1)$

$"Again "sin(x/2)=sqrt(1/2(1-cosx))$

$=sqrt(1/2(1+(2sqrt2)/3))$

$=sqrt(1/6(3+2sqrt2)$

$=sqrt(1/6((sqrt2)^2+1^2+2sqrt2*1)$

$=sqrt(1/6(sqrt2+1)^2$

$=1/sqrt6(sqrt2+1)$

$:.tan(x/2)=(sin(x/2))/(cos(x/2))=(sqrt2+1)/(sqrt2-1)$

$=(sqrt2+1)/(sqrt2-1)*(sqrt2+1)/(sqrt2+1)$

$=(sqrt2+1)^2/(2-1)=3+2sqrt2$