Question

Please solve the question in attachment ​

Answer 1

Here,

$\longrightarrow\sf{A+B+C=\pi}$

$\longrightarrow\sf{B+C=\pi-A}$

Taking tangent on both sides,

$\longrightarrow\sf{\tan(B+C)=\tan(\pi-A)}$

$\longrightarrow\sf{\dfrac{\tan B+\tan C}{1-\tan B\tan C}=-\tan A}$

$\longrightarrow\sf{\tan B+\tan C=\tan A\tan B\tan C-\tan A}$

$\longrightarrow\sf{\tan A+\tan B+\tan C=\tan A\tan B\tan C\quad\quad\dots(1)}$

On squaring both sides,

$\longrightarrow\sf{(\tan A+\tan B+\tan C)^2=(\tan A\tan B\tan C)^2}$

$\displaystyle\longrightarrow\sf{\sum\tan^2A+2\sum\tan A\tan B=\tan^2A\tan^2B\tan^2C}$

$\displaystyle\longrightarrow\sf{\sum\tan^2A=\tan^2A\tan^2B\tan^2C-2\sum\tan A\tan B\quad\quad\dots(2)}$

Thus,

$\displaystyle\longrightarrow\sf{\sum\left(\dfrac{\tan A}{\tan B\tan C}\right)=\sum\left(\dfrac{\tan^2A}{\tan A\tan B\tan C}\right)}$

$\displaystyle\longrightarrow\sf{\sum\left(\dfrac{\tan A}{\tan B\tan C}\right)=\dfrac{\displaystyle\sum\tan^2A}{\tan A\tan B\tan C}}$

From (2),

$\displaystyle\longrightarrow\sf{\sum\left(\dfrac{\tan A}{\tan B\tan C}\right)=\dfrac{\tan^2A\tan^2B\tan^2C-2\displaystyle\sum\tan A\tan B}{\tan A\tan B\tan C}}$

$\displaystyle\longrightarrow\sf{\sum\left(\dfrac{\tan A}{\tan B\tan C}\right)=\tan A\tan B\tan C-\dfrac{2\displaystyle\sum\tan A\tan B}{\tan A\tan B\tan C}}$

$\displaystyle\longrightarrow\sf{\sum\left(\dfrac{\tan A}{\tan B\tan C}\right)=\tan A\tan B\tan C-2\sum\dfrac{\tan A\tan B}{\tan A\tan B\tan C}}$

$\displaystyle\longrightarrow\sf{\sum\left(\dfrac{\tan A}{\tan B\tan C}\right)=\tan A\tan B\tan C-2\sum\dfrac{1}{\tan A}}$

From (1),

$\displaystyle\longrightarrow\sf{\sum\left(\dfrac{\tan A}{\tan B\tan C}\right)=\tan A+\tan +B\tan C-2\sum\cot A}$

Or,

$\displaystyle\longrightarrow\sf{\underline{\underline{\sum\left(\dfrac{\tan A}{\tan B\tan C}\right)=\sum(\tan A)-2\sum(\cot A)}}}$

Hence Proved!