Question

cos9°+sin9°|cos9°-sin9°

Answer 1

heya friend
here is your answer.
cos9°+sin9°/cos9°-sin9°
dividing numerator and denominator by cos9°
=(cos9°+sin9°)/cos9°/ (cos9°-sin9°)/cos9°
=1+tan9°/1-tan9°
=tan45°+tan9°/1 - tan45°×tan9°
=tan(45°+9°)
=tan54°
so the answer is tan54°
mark it as brainliest answer if it is helpful.

Answer 2

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.