Question

Two angles of a triangle are cot inverse 2 and cot inverse 3. The third angle is

Answer 1

Answer:

The third angle is $\frac{3\pi}{4}$

Step-by-step explanation:

Formula used:

$cot^{-1}x=tan^{-1}(\frac{1}{x})$

$tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy})$

Let the three angles of a triangle be

A,B and C

Then,

$A+B+C=\pi$

$cot^{-1}2+cot^{-1}3+C=\pi$

$tan^{-1}(\frac{1}{2})+tan^{-1}(\frac{1}{3})+C=\pi$

$tan^{-1}(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}})+C=\pi$

$tan^{-1}(\frac{\frac{3+2}{6}}{1-\frac{1}{6}})+C=\pi$

$tan^{-1}(\frac{\frac{5}{6}}{\frac{5}{6}})+C=\pi$

$tan^{-1}(1)+C=\pi$

$\frac{\pi}{4}+C=\pi$

$C=\pi-\frac{\pi}{4}$

$C=\frac{3\pi}{4}$