Simplifying $\tan^{-1} {\cot(\frac{-1}4)}$

Question

Simplify $\tan^{-1} {\cot(\frac{-1}4)}$

I have been trying to solve this and I end up with 4, while the correct answer is $(\frac{1}4 - \frac{\pi}{2})$. Please help

Answer 1

Note that

$$\cot \left(-\frac14\right)=-\tan\left(\frac{\pi}{2}-\frac14\right)=\tan\left(\frac14-\frac{\pi}{2}\right)$$

thus

$$\tan^{-1}\cot \left(-\frac14\right)=\tan^{-1}\tan\left(\frac14-\frac{\pi}{2}\right)=\frac14-\frac{\pi}{2}$$

Answer 2

Hint : $\tan({\pi \over 2} - x) ={ \sin({\pi \over 2} - x) \over \cos({\pi \over 2} - x) } = {\cos(x) \over \sin(x)} = {1\over\tan(x)}$