Question

If \[\cot x + \cos ecx = \sqrt 3 \] then find the principal value of \[\left( {x - \dfrac{\pi }{6}} \right)\]
A.\[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi }{4}\]
C. \[\dfrac{\pi }{2}\]
D. \[\dfrac{\pi }{6}\]

Answer 1

Hint: First write the given equation in \[\sin ,\cos \] form, then solve it. After solving the equation write the general value of x as \[x = 2n\pi + \dfrac{\pi }{3}\], then subtract \[\dfrac{\pi }{6}\] from each side to get the form \[\left( {x - \dfrac{\pi }{6}} \right)\]. Then substitute 0 for n in the equation \[x{\rm{ }}-{\rm{ }}\dfrac{\pi }{6}{\rm{ }} = {\rm{ }}2n\pi {\rm{ }} + {\rm{ }}\dfrac{\pi }{6}\] to obtain the principal value.



Formula Used:The general solution of
 \[\begin{array}{l}\tan x = \tan\theta \\ \Rightarrow x = n\pi + \theta \end{array}\]
And \[1 + \cos 2A = 2{\cos ^2}A\],
 \[\sin 2A = 2\sin A\cos A\] .



Complete step by step solution:The given equation is,
\[\cot x{\rm{ }} + {\rm{ }}\cos ecx{\rm{ }} = {\rm{ }}\surd 3\]
\[\dfrac{{\cos x}}{{\sin x}}{\rm{ + }}\dfrac{1}{{\sin x}}{\rm{ }} = {\rm{ }}\surd 3\]
\[\dfrac{{1 + \cos x}}{{\sin x}} = {\rm{ }}\surd 3\]
\[\dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{\rm{ }} = {\rm{ }}\surd 3\]
\[\dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}}{\rm{ }} = {\rm{ }}\surd 3\]
\[{\rm{cot }}\left( {\dfrac{x}{2}} \right){\rm{ }} = {\rm{ }}\surd 3\]
Take the reciprocal of both side of the equation.
Now,
\[\tan {\rm{ }}\left( {\dfrac{x}{2}} \right){\rm{ }} = {\rm{ }}\dfrac{1}{{\sqrt 3 }}\]
\[\dfrac{x}{2}{\rm{ }} = {\rm{ }}n\pi {\rm{ }} + {\rm{ }}\dfrac{\pi }{6}\]
\[x{\rm{ }} = {\rm{ }}2n\pi {\rm{ }} + {\rm{ }}\dfrac{\pi }{3}\]
Subtract \[\dfrac{\pi }{6}\] from both sides of the equation,
\[x{\rm{ }}-{\rm{ }}\dfrac{\pi }{6}{\rm{ }} = {\rm{ }}2n\pi {\rm{ }} + {\rm{ }}\dfrac{\pi }{6}\]
Substitute 0 for n in the equation \[x{\rm{ }}-{\rm{ }}\dfrac{\pi }{6}{\rm{ }} = {\rm{ }}2n\pi {\rm{ }} + {\rm{ }}\dfrac{\pi }{6}\] to obtain the principal value.
Therefore, the principal value is \[\dfrac{\pi }{6}\] .



Option ‘D’ is correct



Note: Sometime students only write \[\tan {\rm{ }}\left( {\dfrac{x}{2}} \right){\rm{ }} = {\rm{ tan }}\dfrac{\pi }{6}\], as \[{\rm{tan }}\dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}\] this concept is correct but for generalisation we need to write all the angles for which tangent takes the value \[\dfrac{1}{{\sqrt 3 }}\], so we need to write it as \[\dfrac{x}{2}{\rm{ }} = {\rm{ }}n\pi {\rm{ }} + {\rm{ }}\dfrac{\pi }{6}\] .