$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{5\pi}{2}$
$\frac{\pi}{6}$
Solution:
We know,
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{a}}\Big)=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{a}}{\text{b}+\text{c}}+\frac{\text{b}}{\text{c}+\text{a}}}{1-\frac{\text{a}}{\text{b}+\text{c}}\times\frac{\text{b}}{\text{c}+\text{a}}}\end{pmatrix}$
$=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{ac}+\text{a}^2+\text{b}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}{\frac{\text{ac}+\text{c}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}\end{pmatrix}$
$=\tan^{-1}\Big(\frac{\text{ac}+\text{c}^2+\text{bc}}{\text{ac}+\text{c}^2+\text{bc}}\Big)$ $\big[\because\text{a}^2+\text{b}^2=\text{c}^2\big]$
$=\tan^{-1}(1)$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ If $1$ ball is drawn from each of the boxes $B_1, B_2$ and $B_3$, the probability that all $3$ drawn balls are of the same colour is
$(A)$ $\frac{82}{648}$ $(B)$ $\frac{90}{648}$ $(C)$ $\frac{558}{648}$ $(D)$ $\frac{566}{648}$
$2.$ If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $2$ balls are drawn from bo $B _2$ is
$(A)$ $\frac{116}{181}$ $(B)$ $\frac{126}{181}$ $(C)$ $\frac{65}{181}$ $(D)$ $\frac{55}{181}$
Give the answer question $1$ and $2.$