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

Concept: Inverse trigonometric identities and the formula for $\cot(A-B)$.

We start by letting $\theta = \cot^{-1}3$, so $\cot\theta = 3$. Then the expression becomes $\cot\left(\frac{\pi}{4} - 2\theta\right)$.

First, find $\cot 2\theta$ using the double-angle formula:

Now apply the $\cot(A-B)$ identity:

Since $\cot\frac{\pi}{4} = 1$, substitute:

Let’s start with the intuition. The problem asks us to prove that a messy-looking trigonometric expression simplifies to the neat integer 7. The core challenge is the nested inverse function: $2\cot^{-1}3$ inside a cotangent of a shifted angle.

Whenever you see $\cot^{-1}$, it’s often easier to convert to $\tan^{-1}$ because the tangent addition/subtraction formulas are more familiar. Remember: $\cot^{-1} x = \tan^{-1}(1/x)$ for $x>0$. Since 3 is positive, we can safely do that.

Then the angle becomes $\frac{\pi}{4} - 2\tan^{-1}(1/3)$. The $\frac{\pi}{4}$ suggests using the tangent subtraction formula: $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. And since we ultimately want the cotangent, we can compute the tangent first and then take its reciprocal.

Let’s work through it step by step.

1. Rewrite the inverse cotangent For $x>0$, $\cot^{-1} x = \tan^{-1}(1/x)$. So:

Hence the given expression becomes:

2. Let $\theta = \tan^{-1}(1/3)$ Then $\tan\theta = \frac{1}{3}$. We need $\tan(2\theta)$ because the angle inside is $\frac{\pi}{4} - 2\theta$. Using the double-angle formula:

3. Now find $\tan\left(\frac{\pi}{4} - 2\theta\right)$ Use the subtraction formula:

4. Convert tangent to cotangent Since $\cot x = \frac{1}{\tan x}$ (provided $\tan x \neq 0$), we have:

Method: Simplifying $\cot$ (or $\tan$) of an expression built from $2\cot^{-1}$

This is the general route for identities like $\cot\!\big(\tfrac{\pi}{4}-2\cot^{-1}a\big)$: name the inverse as an angle, use a double-angle formula, then a compound-angle formula.

Steps

Step 1: Let the inverse be a single angle.

Put $\theta=\cot^{-1}a$, so $\cot\theta=a$. The expression becomes a function of $\theta$ only.

Step 2: Handle the "$2$" with a double-angle identity.

Step 3: Apply the compound-angle formula.

Substitute the known values and simplify the resulting fraction of fractions.

Step 4 (equally valid): the tangent route.

Convert $\cot^{-1}a=\tan^{-1}\tfrac1a$ (for $a>0$), compute $\tan$ of the whole angle with $\tan(A-B)$, then take the reciprocal at the end. Pick whichever keeps the arithmetic cleaner.

Common Mistakes

Mistake 1: Reading $2\cot^{-1}3$ as $\cot^{-1}(2\cdot3)=\cot^{-1}6$.

Why it's wrong: the $2$ multiplies the angle, not the argument. Correct approach: set $\theta=\cot^{-1}3$ and compute $\cot 2\theta$ with the double-angle formula, not $\cot^{-1}6$.

Mistake 2: Using $\cot 2\theta=2\cot\theta$.

Why it's wrong: there is no such linear rule. Correct approach: $\cot 2\theta=\dfrac{\cot^2\theta-1}{2\cot\theta}=\dfrac{9-1}{6}=\dfrac43$.

Mistake 3: Misremembering the sign in $\cot(A-B)$.

Why it's wrong: writing $\cot(A-B)=\dfrac{\cot A\cot B-1}{\cot B+\cot A}$ (wrong signs) breaks the result. Correct approach: the correct form is $\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$, giving $\dfrac{1\cdot\frac43+1}{\frac43-1}=7$.