Trigonometry Simplification | SMO, 2009 | Problem 26

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Trigonometry Simplification.

If \(\frac {cos 100^\circ}{1-4 sin 25^\circ cos 25^\circ cos 50^\circ} = tan x^\circ \)

Find \( x^\circ \) ?

If you really got stuck into this sum we can start from here

\(\frac {cos 100^\circ}{1-4 sin 25^\circ cos 25^\circ cos 50^\circ}\)

= \(\frac {cos 100^\circ}{1-2sin 50^\circ cos 50^\circ}\)

Now let's check with some basic values in trigonometry

\( Cos 2 A = cos^2 A - sin^2 A \) and

\(2 sin A cos A = sin 2 A\)

Now try the rest of the sum by using these two above mentioned values..................

Let's continue from the last hint :

\( cos 100^\circ = cos^2 50^\circ - sin^2 50^\circ \)

\( 2 sin 25^\circ cos 25^\circ = sin 50^\circ\)

\(\frac {cos^2 50^\circ - sin^2 50^\circ}{2sin 50^\circ cos 50^\circ}\)

\(\frac {cos^2 50^\circ - sin^2 50^\circ }{(cos 50^\circ - sin 50^\circ)^2}\)

Using \(a^2 - b^2 = (a+b) (a-b)\) formula

\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ - sin 50^\circ}\)

Do the rest of the steps .................

Starting from right after the last hint:

\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ - sin 50^\circ}\)

= \(\frac {1+ tan 50^\circ}{1-tan 50^\circ}\)

= \(\frac {tan 45^\circ + tan 50^\circ}{1-tan 45^\circ tan 50 ^\circ}\)

= \( tan 95^\circ\) - Answer

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