# If $\dfrac{\tan{\theta}}{\tan{\theta}-\tan{3\theta}} = \dfrac{1}{3}$, then find the value of $\dfrac{\cot{\theta}}{\cot{\theta}-\cot{3\theta}}$

It is given that the value of $\dfrac{\tan{\theta}}{\tan{\theta}-\tan{3\theta}}$ is $\dfrac{1}{3}$ but it is asked to evaluate the value of the trigonometric expression $\dfrac{\cot{\theta}}{\cot{\theta}-\cot{3\theta}}$ in this problem. The value of the trigonometric expression $\dfrac{\cot{\theta}}{\cot{\theta}-\cot{3\theta}}$ can be evaluated on the basis of the given trigonometric equation.

### Simplify the Trigonometric equation

Use cross multiplication method firstly to start simplifying this trigonometric equation.

$\implies$ $3 \times \tan{\theta}$ $\,=\,$ $1 \times (\tan{\theta}-\tan{3\theta})$

$\implies$ $3\tan{\theta}$ $\,=\,$ $\tan{\theta}-\tan{3\theta}$

$\implies$ $3\tan{\theta}-\tan{\theta}$ $\,=\,$ $-\tan{3\theta}$

$\implies$ $2\tan{\theta}$ $\,=\,$ $-\tan{3\theta}$

The term $\tan{(3\theta})$ can be expanded in the trigonometric equation by the tan triple angle formula.

$\implies$ $2\tan{\theta}$ $\,=\,$ $-\Bigg[\dfrac{3\tan{\theta}-\tan^3{\theta}}{1-3\tan^2{\theta}}\Bigg]$

$\implies$ $2\tan{\theta}$ $\,=\,$ $\dfrac{-3\tan{\theta}+\tan^3{\theta}}{1-3\tan^2{\theta}}$

Now, use cross multiplication method one more time to simplify this trigonometric equation further.

$\implies$ $2\tan{\theta}{(1-3\tan^2{\theta})}$ $\,=\,$ $-3\tan{\theta}+\tan^3{\theta}$

$\implies$ $2\tan{\theta} \times 1$ $\,-\,$ $2\tan{\theta} \times 3\tan^2{\theta}$ $\,=\,$ $-3\tan{\theta}+\tan^3{\theta}$

$\implies$ $2\tan{\theta} \,-\, 6\tan^3{\theta}$ $\,=\,$ $-3\tan{\theta}+\tan^3{\theta}$

$\implies$ $2\tan{\theta}$ $+$ $3\tan{\theta}$ $\,=\,$ $6\tan^3{\theta}$ $+$ $\tan^3{\theta}$

$\implies$ $5\tan{\theta}$ $\,=\,$ $7\tan^3{\theta}$

$\implies$ $\dfrac{5}{7}$ $\,=\,$ $\dfrac{\tan^3{\theta}}{\tan{\theta}}$

$\implies$ $\dfrac{\tan^3{\theta}}{\tan{\theta}}$ $\,=\,$ $\dfrac{5}{7}$

$\implies$ $\require{cancel} \dfrac{\cancel{\tan^3{\theta}}}{\cancel{\tan{\theta}}}$ $\,=\,$ $\dfrac{5}{7}$

$\,\,\, \therefore \,\,\,\,\,\, \tan^2{\theta} \,=\, \dfrac{5}{7}$

Finally, the trigonometric equation is successfully simplified and it is evaluated that the square of tangent of angle theta is equal to $5/7$. In this trigonometry problem, no need to evaluate $\tan{\theta}$ value but just remember the value of $\tan^2{\theta}$.

### Simplify the Trigonometric expression

It is required to find the value of the trigonometric expression $\dfrac{\cot{\theta}}{\cot{\theta}-\cot{3\theta}}$. So, let’s simplify this trigonometric expression to evaluate it.

Now, expand the $\cot{(3\theta)}$ term in the denominator of the fraction by using cot triple angle formula.

$= \,\,\,$ $\dfrac{\cot{\theta}}{\cot{\theta}-\dfrac{3\cot{\theta}-\cot^3{\theta}}{1-3\cot^2{\theta}}}$

$\cot{\theta}$ is a common factor in the numerator of the expansion of $\cot{3\theta}$ term. It can be taken common from them to simplify this expression further.

$= \,\,\,$ $\dfrac{\cot{\theta}}{\cot{\theta}-\dfrac{\cot{\theta}{(3-\cot^2{\theta})}}{1-3\cot^2{\theta}}}$

$= \,\,\,$ $\dfrac{\cot{\theta}}{\cot{\theta}-\cot{\theta} \Bigg[ \dfrac{3-\cot^2{\theta}}{1-3\cot^2{\theta}} \Bigg]}$

$= \,\,\,$ $\dfrac{\cot{\theta}}{\cot{\theta} \Bigg[1-\dfrac{3-\cot^2{\theta}}{1-3\cot^2{\theta}}\Bigg]}$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\cot{\theta}}}{\cancel{\cot{\theta}} \Bigg[1-\dfrac{3-\cot^2{\theta}}{1-3\cot^2{\theta}}\Bigg]}$

$= \,\,\,$ $\require{cancel} \dfrac{1}{1-\dfrac{3-\cot^2{\theta}}{1-3\cot^2{\theta}}}$

$= \,\,\,$ $\dfrac{1}{\dfrac{1 \times {(1-3\cot^2{\theta})} \,-\, (3-\cot^2{\theta})}{1-3\cot^2{\theta}}}$

$= \,\,\,$ $\dfrac{1-3\cot^2{\theta}}{1 \times {(1-3\cot^2{\theta})} \,-\, (3-\cot^2{\theta})}$

$= \,\,\,$ $\dfrac{1-3\cot^2{\theta}}{1-3\cot^2{\theta}-3+\cot^2{\theta}}$

$= \,\,\,$ $\dfrac{1-3\cot^2{\theta}}{1-3-3\cot^2{\theta}+\cot^2{\theta}}$

$= \,\,\,$ $\dfrac{1-3\cot^2{\theta}}{-2-2\cot^2{\theta}}$

$= \,\,\,$ $\dfrac{1-3\cot^2{\theta}}{-2-2\cot^2{\theta}}$

It is derived that the value $\tan^2{\theta}$ is $\dfrac{5}{7}$ but the trigonometric expression is in terms of $\cot^2{\theta}$. Actually, the tangent and cotangent functions have direct relation as per reciprocal identity of tangent function. So, the square of cot function can be written as the reciprocal of square of tan function.

$= \,\,\,$ $\dfrac{1-3 \times \dfrac{1}{\tan^2{\theta}}}{-2-2 \times \dfrac{1}{\tan^2{\theta}}}$

### Find the value of the trigonometric expression

It is time to find the value of the trigonometric expression by substituting the value of $tan^2{\theta}$.

$= \,\,\,$ $\dfrac{1-3 \times \dfrac{1}{\dfrac{5}{7}}}{-2-2 \times \dfrac{1}{\dfrac{5}{7}}}$

$= \,\,\,$ $\dfrac{1-3 \times \dfrac{7}{5}}{-2-2 \times \dfrac{7}{5}}$

$= \,\,\,$ $\dfrac{1-\dfrac{3 \times 7}{5}}{-2-\dfrac{2 \times 7}{5}}$

$= \,\,\,$ $\dfrac{1-\dfrac{21}{5}}{-2-\dfrac{14}{5}}$

$= \,\,\,$ $\dfrac{\dfrac{1 \times 5 -21}{5}}{\dfrac{-2 \times 5 -14}{5}}$

$= \,\,\,$ $\dfrac{\dfrac{5-21}{5}}{\dfrac{-10-14}{5}}$

$= \,\,\,$ $\dfrac{\dfrac{-16}{5}}{\dfrac{-24}{5}}$

$= \,\,\,$ $\dfrac{-16}{5} \times \dfrac{5}{-24}$

$= \,\,\,$ $\dfrac{-16}{-24} \times \dfrac{5}{5}$

$= \,\,\,$ $\require{cancel} \dfrac{16}{24} \times \dfrac{\cancel{5}}{\cancel{5}}$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{16}}{\cancel{24}} \times 1$

$= \,\,\,$ $\dfrac{2}{3} \times 1$

$\therefore \,\,\,\,\,\, \dfrac{\cot{\theta}}{\cot{\theta}-\cot{3\theta}}$ $\,=\,$ $\dfrac{2}{3}$

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