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Question: If we have an expression \[x + y + z = 0,\left| x \right| = \left| y \right| = \left| z \right| = 2\...

If we have an expression x+y+z=0,x=y=z=2x + y + z = 0,\left| x \right| = \left| y \right| = \left| z \right| = 2 and θ\theta is angle between yy and zz then the value of 2cosec2θ+3cot2θ2\cos e{c^2}\theta + 3{\cot ^2}\theta is:
1. 113\dfrac{{11}}{3}
2. 83\dfrac{8}{3}
3. 53\dfrac{5}{3}
4. 11

Explanation

Solution

Here in this question we will use the concepts of vectors like mod of vector and dot product of two vectors to find the required solution. Also we have to use the basic trigonometric values in terms of their relation with the sides of a right angled triangle.
A vector is a quantity that has both magnitude, as well as direction. Some mathematical operations can be performed on vectors such as addition and multiplication.

Complete step-by-step solution:
We are given x+y+z=0,x=y=z=2x + y + z = 0,\left| x \right| = \left| y \right| = \left| z \right| = 2 and θ\theta is the angle between yy and zz .
We are to find the value of 2cosec2θ+3cot2θ2\cos e{c^2}\theta + 3{\cot ^2}\theta .
Therefore we get y+z=xy + z = - x
Taking mod on both the sides we get ,
y+z=x\left| {y + z} \right| = \left| x \right|
Squaring both sides we get ,
y+z2=x2{\left| {y + z} \right|^2} = {\left| x \right|^2}
y2+z2+2yzcosθ=x2{\left| y \right|^2} + {\left| z \right|^2} + 2\left| y \right|\left| z \right|\cos \theta = {\left| x \right|^2}
Since we are given x=y=z=2\left| x \right| = \left| y \right| = \left| z \right| = 2
Putting these values in the above equation we get ,
4+4+2(2)(2)cosθ=44 + 4 + 2(2)(2)\cos \theta = 4
On simplifying the expression we get ,
cosθ=12\cos \theta = - \dfrac{1}{2}
Squaring both sides we get ,
cos2θ=14{\cos ^2}\theta = \dfrac{1}{4}
Using the identity sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1.
Hence we get sin2θ=34{\sin ^2}\theta = \dfrac{3}{4}
Therefore consider 2cosec2θ+3cot2θ2\cos e{c^2}\theta + 3{\cot ^2}\theta
This can be rewritten as 2+3cos2θsin2θ\dfrac{{2 + 3{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}
Putting the values of cos2θ{\cos ^2}\theta and sin2θ{\sin ^2}\theta we get ,
2+3cos2θsin2θ=2+3(14)(34)=113\dfrac{{2 + 3{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} = \dfrac{{2 + 3\left( {\dfrac{1}{4}} \right)}}{{\left( {\dfrac{3}{4}} \right)}} = \dfrac{{11}}{3}
Hence we get the value for the required expression.
Therefore option (1) is the correct answer.

Note: To solve such types of questions one needs to have knowledge of vectors and their properties like dot product, cross product etc. Also one must be aware of all the concepts of trigonometry , its related formulas and basic trigonometric identities. In addition to these we must know their relation with the sides of a right angled triangle. We must do the calculations carefully and should recheck at each and every step in order to get the exact solution.