Question

prove that for all x ∈ R . cot^-1(-x)=π-cot^-1(x)​

Answer 1

Answer:

Step-by-step explanation:

$\tt{Let\,\,\,cot^{-1}(-x)=\theta\,\,,\,\,\,\theta\in(0,\pi)}$

$\tt{\implies\,-x=cot(\theta)\,\,,\,\,\,\theta\in(0,\pi)}$

$\tt{\implies\,x=-cot(\theta)}$

Now, we know,

$\bf{cot(\pi-\phi)=-cos(\phi)\,\,\,\,\,\,and\,\,\,\,\,\,cot(2\pi-\phi)=-cot(\phi)}$

But only $\pi-\phi \in (0,\pi)$

So,

$\tt{\implies\,x=cot(\pi-\theta)}$

$\tt{\implies\,cot^{-1}(x)=\pi-\theta}$

$\tt{\implies\,\theta=\pi-cot^{-1}(x)}$

Hence,

$\tt{\implies\,cot^{-1}(-x)=\pi-cot^{-1}(x)}$