Question

If $\cot \theta + \cot \left( {\dfrac{\pi }{4} + \theta } \right) = 2$, then the general value of $\theta$ is:
A. $2n\pi \pm \dfrac{\pi }{6}$
B. $2n\pi \pm \dfrac{\pi }{3}$
C. $n\pi \pm \dfrac{\pi }{3}$
D. $n\pi \pm \dfrac{\pi }{6}$

Answer 1

The given question involves solving a trigonometric equation and finding the general value of angle $\theta$ that satisfies the given equation and lies in the range of $[0,2\pi ]$. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities. We will use the compound angle formula of cotangent $\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$ and then simplify the equation to find the general values for the angle $\theta$.

Complete step by step answer:
In the given problem, we have to solve the trigonometric equation $\cot \theta + \cot \left( {\dfrac{\pi }{4} + \theta } \right) = 2$ and find the values of $\theta$ that satisfy the given equation.
So, In order to solve the given trigonometric equation$\cot \theta + \cot \left( {\dfrac{\pi }{4} + \theta } \right) = 2$ , we will open up the cotangent of the sum of two angles using the compound angle trigonometric formula $\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$.
So, we get, $\cot \theta + \cot \left( {\dfrac{\pi }{4} + \theta } \right) = 2$
$\Rightarrow \cot \theta + \dfrac{{\cot \left( {\dfrac{\pi }{4}} \right)\cot \theta - 1}}{{\cot \left( {\dfrac{\pi }{4}} \right) + \cot \theta }} = 2$
Now, we know that the value of $\cot \left( {\dfrac{\pi }{4}} \right)$ is one. So, we get,
$\Rightarrow \cot \theta + \dfrac{{\left( 1 \right)\cot \theta - 1}}{{\left( 1 \right) + \cot \theta }} = 2$

Now, taking LCM of both the rational trigonometric expressions, we get,
$\Rightarrow \dfrac{{\cot \theta \left( {1 + \cot \theta } \right) + \cot \theta - 1}}{{1 + \cot \theta }} = 2$
Opening brackets and simplifying the equation, we get,
$\Rightarrow \dfrac{{\cot \theta + {{\cot }^2}\theta + \cot \theta - 1}}{{1 + \cot \theta }} = 2$
Cross multiplying the terms of the equation, we get,
$\Rightarrow {\cot ^2}\theta + 2\cot \theta - 1 = 2\left( {1 + \cot \theta } \right)$
$\Rightarrow {\cot ^2}\theta + 2\cot \theta - 1 = 2 + 2\cot \theta$
Shifting all the terms to the left side of the equation, we get,
$\Rightarrow {\cot ^2}\theta + 2\cot \theta - 2\cot \theta - 2 - 1 = 0$
Cancelling the like terms with opposite signs, we get,
$\Rightarrow {\cot ^2}\theta - 3 = 0$

Now, we know that tangent and cotangent are reciprocal functions of each other. So, we get,
$\Rightarrow \dfrac{1}{{{{\tan }^2}\theta }} - 3 = 0$
Shifting constant term to right side of the equation, we get,
$\Rightarrow \dfrac{1}{{{{\tan }^2}\theta }} = 3$
$\Rightarrow {\tan ^2}\theta = \dfrac{1}{3}$
Now, we know the double angle formula of cosine $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$. So, we get,
$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$
Substituting the value of ${\tan ^2}\theta$, we have,
$\Rightarrow \cos 2\theta = \dfrac{{1 - \left( {\dfrac{1}{3}} \right)}}{{1 + \left( {\dfrac{1}{3}} \right)}}$
$\Rightarrow \cos 2\theta = \dfrac{{\left( {\dfrac{2}{3}} \right)}}{{\left( {\dfrac{4}{3}} \right)}}$
Simplifying the expression,
$\Rightarrow \cos 2\theta = \dfrac{1}{2}$

Now, we know that the value of $\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}$. So, we get,
$\Rightarrow \cos 2\theta = \cos \dfrac{\pi }{3}$
Now, we know that the general solution of equation $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$.
Hence, we get,
$\Rightarrow 2\theta = 2n\pi \pm \left( {\dfrac{\pi }{3}} \right)$
On dividing both sides by $2$, we get,
$\therefore \theta = n\pi \pm \left( {\dfrac{\pi }{6}} \right)$

So, the correct answer is option D.

Note: The given trigonometric equation can also be solved by first converting the equation into tangent form using the trigonometric formula $\tan x = \dfrac{1}{{\cot x}}$ and then solving further. We should remember the double angle formula of cosine and the compound angle formula for cotangent to solve the given question. One must have good grip over simplification rules in order to get to the required answer.