Q.If $\tan(\pi\cos\theta) = \cot(\pi\sin\theta)$, then $\cos\left(\theta - \dfrac{\pi}{4}\right) = \pm\dfrac{1}{2\sqrt{2}}$.

The key idea is to use trigonometric identities to simplify the given equation and then apply the general solution for tangent functions.

Given $\tan(\pi\cos\theta) = \cot(\pi\sin\theta)$.

1. Use the identity $\cot x = \tan\left(\dfrac{\pi}{2} - x\right)$ to rewrite the equation: $\tan(\pi\cos\theta) = \tan\left(\dfrac{\pi}{2} - \pi\sin\theta\right)$
2. For $\tan A = \tan B$, the general solution is $A = n\pi + B$, where $n$ is an integer. $\pi\cos\theta = n\pi + \dfrac{\pi}{2} - \pi\sin\theta$
3. Divide by $\pi$ and rearrange terms: $\cos\theta = n + \dfrac{1}{2} - \sin\theta$ $\cos\theta + \sin\theta = n + \dfrac{1}{2}$ Since $-\sqrt{2} \le \cos\theta + \sin\theta \le \sqrt{2}$, we have $-\sqrt{2} \le n + \dfrac{1}{2} \le \sqrt{2}$. This implies $-1.414 \le n + 0.5 \le 1.414$, so $-1.914 \le n \le 0.914$. The only integer values for $n$ are $0$ and $-1$. If $n=0$, $\cos\theta + \sin\theta = \dfrac{1}{2}$. If $n=-1$, $\cos\theta + \sin\theta = -1 + \dfrac{1}{2} = -\dfrac{1}{2}$. Thus, $\cos\theta + \sin\theta = \pm\dfrac{1}{2}$.
4. Now, consider the expression $\cos\left(\theta - \dfrac{\pi}{4}\right)$. Using the compound angle formula $\cos(A-B) = \cos A \cos B + \sin A \sin B$: …

When we encounter an equation involving different trigonometric functions, like $\tan$ and $\cot$ here, the first step is usually to express them in terms of the same function. This allows us to use general solutions for trigonometric equations. The identity $\cot x = \tan(\pi/2 - x)$ is key for this.

Once we have an equation of the form $\tan A = \tan B$, its general solution is $A = n\pi + B$, where $n$ is an integer. This step introduces an integer $n$, which we must then constrain using the known bounds of trigonometric expressions.

Finally, the expression we need to find, $\cos(\theta - \pi/4)$, is a specific linear combination of $\cos\theta$ and $\sin\theta$. Recognizing that $\cos\theta + \sin\theta$ can be rewritten as $\sqrt{2}\cos(\theta - \pi/4)$ is crucial for connecting the derived equation to the target expression.

Let's break down the solution:

1. Convert $\cot$ to $\tan$:

   The given equation is $\tan(\pi\cos\theta) = \cot(\pi\sin\theta)$.

   We use the identity $\cot x = \tan\left(\dfrac{\pi}{2} - x\right)$.

   Applying this to the right side, we get:

   $\tan(\pi\cos\theta) = \tan\left(\dfrac{\pi}{2} - \pi\sin\theta\right)$

2. Apply the general solution for $\tan A = \tan B$:

   If $\tan A = \tan B$, then $A = n\pi + B$, where $n$ is an integer.

   Here, $A = \pi\cos\theta$ and $B = \dfrac{\pi}{2} - \pi\sin\theta$.

   So, we have:

   $\pi\cos\theta = n\pi + \left(\dfrac{\pi}{2} - \pi\sin\theta\right)$

3. Simplify the equation:

   Divide the entire equation by $\pi$:

   $\cos\theta = n + \dfrac{1}{2} - \sin\theta$

   Rearrange the terms to group $\cos\theta$ and $\sin\theta$:

   $\cos\theta + \sin\theta = n + \dfrac{1}{2}$

4. Constrain the integer $n$ using the range of $\cos\theta + \sin\theta$:

   We know that for any real $\theta$, the expression $\cos\theta + \sin\theta$ has a specific range.

   For $a\cos x + b\sin x$, the range is $\left[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}\right]$.

   For $\cos\theta + \sin\theta$, we have $a=1$ and $b=1$.

   So, the range of $\cos\theta + \sin\theta$ is $\left[-\sqrt{1^2+1^2}, \sqrt{1^2+1^2}\right] = [-\sqrt{2}, \sqrt{2}]$.

   Therefore, we must have:

   $-\sqrt{2} \le n + \dfrac{1}{2} \le \sqrt{2}$

   Numerically, $\sqrt{2} \approx 1.414$.

   $-1.414 \le n + 0.5 \le 1.414$

   Subtract $0.5$ from all parts of the inequality:

   $-1.414 - 0.5 \le n \le 1.414 - 0.5$

   $-1.914 \le n \le 0.914$

   Since $n$ must be an integer, the possible values for $n$ are $-1$ and $0$.

5. Relate $\cos\theta + \sin\theta$ to $\cos\left(\theta - \dfrac{\pi}{4}\right)$:

   We want to express $\cos\theta + \sin\theta$ in the form of $\cos\left(\theta - \dfrac{\pi}{4}\right)$.

   Recall the cosine subtraction formula: $\cos(A-B) = \cos A \cos B + \sin A \sin B$.

   Let $A = \theta$ and $B = \dfrac{\pi}{4}$.

   $\cos\left(\theta - \dfrac{\pi}{4}\right) = \cos\theta \cos\left(\dfrac{\pi}{4}\right) + \sin\theta \sin\left(\dfrac{\pi}{4}\right)$ …