Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry.
Problem on Trigonometry | SMO, 2008 |Problem 22
Find the value of \(\frac {tan 40^\circ tan 60^\circ tan 80^\circ}{tan40^\circ + tan 60^\circ + tan 80^\circ}\)
- 1
- 15
- 6
- 0
Key Concepts
Trigonometry
Tan Rule
Check the Answer
But try the problem first…
Answer: 1
Singapore Mathematical Olympiad, 2008
Challenges and Thrills – Pre College Mathematics
Try with Hints
First hint
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………………
Second Hint
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……………..
Final Step
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……………….
AMC – AIME Boot Camp… for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad