95points! Prove (2cos12°cos6°)/(1+2sin12°cos6°)=tan54°
Answers
Answer:
LHS = (Cos9° + Sin9°) / (Cos9° - Sin9°)
Divide by Nr and Dr by Cos9° ...
LHS = (1 + Tan9°) / (1 - Tan9°)
= (Tan 45° + Tan 9°) / (1 - Tan 45° * Tan 9°)
= Tan (45°+9°)
= Tan 54°
= Cot 36°.
There could be many forms in which the result could be expressed. I dont know which one you are looking for...
LHS = (cos 9° + sin 9°)/(cos 9° - Sin 9°)
Multiply Nr and Dr. with Cos9° + sin 9° .
LHS = (cos 9° + sin9°)² / (cos ² 9° - sin² 9°)
= (1 + 2 sin 9° * Cos 9°) / Cos (2*9°)
= (1+sin18)/ Cos 18°
= Sec 18° + Tan 18°
We find the values of Sin18° and Cos18° as follows:
We know : Cos 54° = Sin 36° . Let A = 18°.
Sin 2A = 2 Sin A Cos A.
Cos 3A = Cos A (4 Cos² A - 3) = Cos A (1 - 4 Sin²A)
=> Cos 3A = Sin 2A
Cos A (1 - 4 SIn² A) = 2 Sin A Cos A
So 4 Sin² A + 2 Sin A - 1 = 0
Sin 18° = (√5 - 1)/4
Using Cos² 18° = 1 - Sin² 18°, we get: Cos 18° = √[10 + 2√5 ]/4
Now LHS = (1+Sin 18°)/Cos 18°
= (3 + √5) / √[10 + 2√5]
This is the answer as an irrational number.