Question

Tan inverse1 + cos inverse-1/2

Answer 1

${tan}^{ - 1} 1 + {cos}^{ - 1} ( - \frac{1}{2} ) \\ let \: \: \: tan ^{ - 1} 1 = x \\ \to \: tanx = 1 \\ \to tanx = tan \frac{\pi}{4} \\ x = \frac{\pi}{4} \\ we \: know \: range \: of \: {tan}^{ - 1} is \: ( - \frac{\pi}{2} \: , \: \frac{\pi}{2} ) \\ so\: pricipal \: value \: of \: tan ^{ - 1} 1 \: is \: \frac{\pi}{4} \\ let \: \: \: \: y = {cos}^{ - 1} ( - \frac{1}{2} ) \\ \to \: cosy = - \frac{1}{2} \\ \to \: cosy = - cos \frac{\pi}{3} \\ \to \: cosy = cos(\pi - \frac{\pi}{3} ) \\ \to \: cosy = cos \frac{2\pi}{3} \\ y = \frac{2\pi}{3} \\ range \: of \: {cos}^{ - 1} \: is \: [0 \: ,\: \pi] \\ so \: \: \: {tan}^{ - 1} 1 + {cos}^{ - 1} ( - \frac{1}{2} ) \\ = x + y \\ = \frac{\pi}{4} + \frac{2\pi}{3} \\ = \frac{11\pi}{12}$