Problem on Trigonometry | SMO, 2008 | Problem - 22

If you got stuck in this sum how to get started you can start by consider a general case where \(40^\circ = A\) , \(60^\circ = B\) and \(80^\circ = C\).

So , A+B+C = \( 180 ^\circ\)

\( A+B = 180^\circ - C\)

(tan (A+B) = tan \(180^\circ - C)\)............................(1)

Now try to implement the basic formula and try to do this sum..................

In this we can continue from the last hint:

the formula of tan (A + B) = \(\frac {tan A + tan B}{1- tan A . tan B}\)

From the equation (1) .......

tan (A+B) = tan (180 - C)

\(\frac {tan A + tan B}{1- tan A . tan B} = tan (180^\circ - c)\)

(frac {tan A + tan B}{1- tan A . tan B} = -tan C )

Now just rearrange this expression and you will get the final answer.................

Here is the final solution:

{tan A + tan B} = -tan C {1- tan A . tan B}

tan A + tan B = - tan C + tan C tan A tan B

tan A + tan B + tan C = tan A tan B tan C

\(\frac {tan A tan B tan C}{tan A + tan B + tan C} = 1\)

Which is the given question. It can be a proof also...................