Question

# Prove that $\cot \left( {\dfrac{\pi }{4} - 2{{\cot }^{ - 1}}3} \right) = 7$

Hint: We solve this question by taking ${\cot ^{ - 1}}3 = A$ and substitute in the equation and then convert the value into $\tan$ form as $\cot \theta = \dfrac{1}{{\tan \theta }}$, apply formula for $\tan (a - b) = \dfrac{{\tan (a) - \tan (b)}}{{1 + \tan (a)\tan (b)}},\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$ and solve the equation.

Let us assume the value ${\cot ^{ - 1}}3 = A$
Therefore taking $\cot$ on both sides of the equation we get
$\cot ({\cot ^{ - 1}}3) = \cot A$
Since function and its inverse cancel out, therefore we get
$3 = \cot A$
Consider the LHS of the equation and substitute ${\cot ^{ - 1}}3 = A$ in it.
$\cot \left( {\dfrac{\pi }{4} - 2{{\cot }^{ - 1}}3} \right) = \cot \left( {\dfrac{\pi }{4} - 2A} \right)$
Since we know, $\cot \theta = \dfrac{1}{{\tan \theta }}$
$\Rightarrow \cot \left( {\dfrac{\pi }{4} - 2A} \right) = \dfrac{1}{{\tan \left( {\dfrac{\pi }{4} - 2A} \right)}}$
Using the formula $\tan (a - b) = \dfrac{{\tan (a) - \tan (b)}}{{1 + \tan (a)\tan (b)}},a = \dfrac{\pi }{4},b = 2A$ in the denominator of the equation

$= \dfrac{1}{{\left( {\dfrac{{\tan \dfrac{\pi }{4} - \tan 2A}}{{1 + \tan \dfrac{\pi }{4}\tan 2A}}} \right)}}$
Multiplying both numerator and denominator by ${{1 + \tan \dfrac{\pi }{4}\tan 2A}}$
$= \dfrac{1}{{\left( {\dfrac{{\tan \dfrac{\pi }{4} - \tan 2A}}{{1 + \tan \dfrac{\pi }{4}\tan 2A}}} \right)}} \times \dfrac{{1 + \tan \dfrac{\pi }{4}\tan 2A}}{{1 + \tan \dfrac{\pi }{4}\tan 2A}}$
Cancel out the same terms from numerator and denominator.
$= \dfrac{{1 + \tan \dfrac{\pi }{4}\tan 2A}}{{\tan \dfrac{\pi }{4} - \tan 2A}}$
Substitute the value of $\tan \dfrac{\pi }{4} = 1$
$= \dfrac{{1 + 1 \times \tan 2A}}{{1 - \tan 2A}} \\ = \dfrac{{1 + \tan 2A}}{{1 - \tan 2A}} \\$ … (i)
Now we calculate the value of $\tan 2A$
Now since we know $\cot \theta = \dfrac{1}{{\tan \theta }}$
Substituting the value of $\theta = A$ we get
$\cot A = \dfrac{1}{{\tan A}} \\ 3 = \dfrac{1}{{\tan A}} \\$
Taking reciprocal on both sides we get
$\tan A = \dfrac{1}{3}$
From the formula $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$, substitute $x = A$ in the formula
$\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
Now substitute the value of $\tan A = \dfrac{1}{3}$ in the equation.
$\tan 2A = \dfrac{{2 \times \dfrac{1}{3}}}{{1 - {{\left( {\dfrac{1}{3}} \right)}^2}}}$
$= \dfrac{{\dfrac{2}{3}}}{{1 - \dfrac{1}{9}}} \\ = \dfrac{{\dfrac{2}{3}}}{{\dfrac{8}{9}}} \\$
Writing the fraction in form of multiplication
$= \dfrac{2}{3} \times \dfrac{9}{8}$
Cancel out the factors from numerator and denominator
$= \dfrac{3}{4}$
So, we get $\tan 2A = \dfrac{3}{4}$
Substitute the value of $\tan 2A = \dfrac{3}{4}$ in equation (i)
$\Rightarrow \dfrac{{1 + \tan 2A}}{{1 - \tan 2A}} = \dfrac{{1 + \dfrac{3}{4}}}{{1 - \dfrac{3}{4}}}$
Take LCM in both numerator and denominator.
$= \dfrac{{\dfrac{{4 + 3}}{4}}}{{\dfrac{{4 - 3}}{4}}} \\ = \dfrac{{\dfrac{7}{4}}}{{\dfrac{1}{4}}} \\$
Writing the fraction in form of multiplication

$= \dfrac{7}{4} \times \dfrac{4}{1}$
Cancel out the common factors from numerator and denominator
$= 7$
Therefore, we get value equal to RHS.
Hence Proved.

Note: Students are likely to make mistakes when they don’t convert the cot value into tan value and start with the solution, since we are familiar with the tan identities so we should convert the values in the beginning from cot to tan. Also, some students try to input the inverse values using calculator or the web which is wrong because we have to prove the statement using properties and not by calculator.