Question

Find the value of \[\cot [\dfrac{\pi }{4} - 2{\cot ^{ - 1}}3]\].
A) $3$
B) $7$
C) $9$
D) $\dfrac{3}{4}$

Answer 1

Hint:
Make a substitution for ${\cot ^{ - 1}}3$, say $x$. Then we can express $\cot $ in terms of $\tan $. After that apply the results of sum formula and duplicate formula of $\tan $, $\tan (A + B)$ and $\tan 2A$ respectively. Simplifying we get the answer.

Useful formula:
We have the following results in trigonometry.
$\cot \theta = \dfrac{1}{{\tan \theta }}$
$\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
$\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$

Complete step by step solution:
We have to find \[\cot [\dfrac{\pi }{4} - 2{\cot ^{ - 1}}3]\].
Let ${\cot ^{ - 1}}3 = x$
This gives, \[\cot [\dfrac{\pi }{4} - 2{\cot ^{ - 1}}3] = \cot [\dfrac{\pi }{4} - 2x]\]
We know that, $\cot \theta = \dfrac{1}{{\tan \theta }}$.
 So, \[\cot [\dfrac{\pi }{4} - 2x] = \dfrac{1}{{\tan [\dfrac{\pi }{4} - 2x]}}\]
Substituting in the above equation we get,
\[\cot [\dfrac{\pi }{4} - 2{\cot ^{ - 1}}3] = \dfrac{1}{{\tan [\dfrac{\pi }{4} - 2x]}} - - - - (i)\]
We have, $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
So, \[\tan [\dfrac{\pi }{4} - 2x] = \dfrac{{\tan \dfrac{\pi }{4} - \tan 2x}}{{1 + \tan \dfrac{\pi }{4}\tan 2x}}\]
Substituting $\tan \dfrac{\pi }{4} = 1$ in the above equation we get,
\[\tan [\dfrac{\pi }{4} - 2x] = \dfrac{{1 - \tan 2x}}{{1 + \tan 2x}}\]
This gives, from ($(i)$,
\[\cot [\dfrac{\pi }{4} - 2{\cot ^{ - 1}}3] = \dfrac{1}{{\dfrac{{1 - \tan 2x}}{{1 + \tan 2x}}}}\]
\[ \Rightarrow \cot [\dfrac{\pi }{4} - 2{\cot ^{ - 1}}3] = \dfrac{{1 + \tan 2x}}{{1 - \tan 2x}} - - - - (ii)\]
Now ${\cot ^{ - 1}}3 = x \Rightarrow \cot x = 3$
So we have, $\tan x = \dfrac{1}{{\cot x}} = \dfrac{1}{3}$
We know, $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
This gives,
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
Substituting we get,
$\tan 2x = \dfrac{{2 \times \dfrac{1}{3}}}{{1 - {{\dfrac{1}{3}}^2}}}$
Simplifying we get,
$\tan 2x = \dfrac{{\dfrac{2}{3}}}{{1 - \dfrac{1}{9}}} = \dfrac{{\dfrac{2}{3}}}{{\dfrac{8}{9}}}$
$ \Rightarrow \tan 2x = = \dfrac{{2 \times 9}}{{8 \times 3}} = \dfrac{3}{4}$
Substituting in $(ii)$ we have,
\[ \Rightarrow \cot [\dfrac{\pi }{4} - 2{\cot ^{ - 1}}3] = \dfrac{{1 + \dfrac{3}{4}}}{{1 - \dfrac{3}{4}}}\]
Simplifying the above equation we get,
\[\cot [\dfrac{\pi }{4} - 2{\cot ^{ - 1}}3] = \dfrac{{\dfrac{7}{4}}}{{\dfrac{1}{4}}}\]
\[ \Rightarrow \cot [\dfrac{\pi }{4} - 2{\cot ^{ - 1}}3] = 7\]

Therefore the answer is option B.

Additional information:
There are many results in trigonometry. Applications of trigonometry are very well found in day to day life. It is mainly used for calculating distances like astronomical distances, which are otherwise difficult to measure.
Trigonometric ratios are the ratios between the sides of a right angled triangle. It can also be represented using the unit circle centred at the origin in the plane.
Trigonometric functions can be made injective by restricting the domain and so can be made invertible. Thus inverse trigonometric ratios also exist. In this question we had seen ${\cot ^{ - 1}}x$.

Note:
The question looks difficult at first view. But introducing the new variable makes it easier. This is the important step here. Then we have to use appropriate formulas in each step.