Question

Please answer the following
#5

Answer 1

All doubts regarding this are welcomed and yes,
All the best for exam :)

Answer 2

Answer   is   (3)..              Follow the simplification step by step...

sin 18° = a
$\sqrt{1-a^2}=cos18^0\\\\\frac{1+cos18^0}{2}=cos^2\frac{18}{2}=cos^2 9^0\\\\\frac{1-cos18^0}{2}=sin^2\frac{18^0}{2}=sin9^0\\\\cos9^0=sin(90-9)=cos81^0\\\\sin9^0=cos(90-0)=cos81^0$

========================  ANOTHER SIMPLE answer...
Cos 81° + Sin 81°
  = Cos 81° + Cos(90° - 81°)
  = Cos 81°  + Cos 9°            ,,, Cos A+ Cos B = 2 cos (A+B/2) cos(A-B)/2

  = 2 Cos (81°+9°)/2  Cos (81°-9°)/2    
  = 2 cos 45° Cos 36
  = 2 *1/√2  *  (1 - 2 Sin² 18°)             as    cos 2A = 1 - 2 sin² A
  = √2 * (1 - 2a²)

This is a very simple form (CORRECT) of the answer.
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