Question

$\tan (\text{A}+\text{B}+\text{C})$ is equal to

• A. $\frac{\tan \text{A}+\tan \text{B}+\tan \text{C}+\tan \text{A}\tan \text{B}\tan \text{C}}{1+\tan \text{A}\tan \text{B}+\tan \text{B}\tan \text{C}+\tan \text{A}\tan \text{C}}$
• B. $\frac{\tan \text{A}+\tan \text{B}+\tan \text{C}+\tan \text{A}\tan \text{B}\tan \text{C}}{1-\tan \text{A}\tan \text{B}-\tan \text{B}\tan \text{C}-\tan \text{A}\tan \text{C}}$
• C. $\frac{\tan \text{A}+\tan \text{B}+\tan \text{C}-\tan \text{A}\tan \text{B}\tan \text{C}}{1-\tan \text{A}\tan \text{B}-\tan \text{B}\tan \text{C}-\tan \text{A}\tan \text{C}}$
• D. $\frac{\tan \text{A}+\tan \text{B}+\tan \text{C}-\tan \text{A}\tan \text{B}\tan \text{C}}{1+\tan \text{A}\tan \text{B}+\tan \text{B}\tan \text{C}+\tan \text{A}\tan \text{C}}$

Answer 1

The correct option is C $\frac{\tan \text{A}+\tan \text{B}+\tan \text{C}-\tan \text{A}\tan \text{B}\tan \text{C}}{1-\tan \text{A}\tan \text{B}-\tan \text{B}\tan \text{C}-\tan \text{A}\tan \text{C}}$

Given : $\tan (\text{A}+\text{B}+\text{C})$

$\tan (\text{A}+\text{B}+\text{C})=\frac{\sin (\text{A}+\text{B}+\text{C})}{\cos (\text{A}+\text{B}+\text{C})}$

We have, $\sin (\text{A}+\text{B}+\text{C})=\cos \text{A}\cos \text{B}\cos \text{C}[\tan \text{A}+\tan \text{B}+\tan \text{C}-\tan \text{A}\tan \text{B}\tan \text{C}]$
$\cos (\text{A}+\text{B}+\text{C})=\cos \text{A}\cos \text{B}\cos \text{C}[1-\tan \text{A}\tan \text{B}-\tan \text{B}\tan \text{C}-\tan \text{A}\tan \text{C}]$

$\therefore \tan (\text{A}+\text{B}+\text{C})=\frac{\cos \text{A}\cos \text{B}\cos \text{C}[\tan \text{A}+\tan \text{B}+\tan \text{C}-\tan \text{A}\tan \text{B}\tan \text{C}]}{\cos \text{A}\cos \text{B}\cos \text{C}[1-\tan \text{A}\tan \text{B}-\tan \text{B}\tan \text{C}-\tan \text{A}\tan \text{C}]}$

$\Rightarrow \tan (\text{A}+\text{B}+\text{C})=\frac{\tan \text{A}+\tan \text{B}+\tan \text{C}-\tan \text{A}\tan \text{B}\tan \text{C}}{1-\tan \text{A}\tan \text{B}-\tan \text{B}\tan \text{C}-\tan \text{A}\tan \text{C}}$