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Exercise - 8.3

Question-1 :-  Evaluate: (i) sin 18°/cos 72°

Solution :-
  We know : cos A = sin (90° – A) 
  So, cos 72° = sin (90° – 72°) = sin 18°
  i.e., sin 18°/cos 72°  = sin 18°/sin 18° = 1
    

(ii)  tan 26°/cot 64°

Solution :-
  We know : cot A = tan (90° – A) 
  So, cot 64° = tan (90° – 64°) = tan 26°
  i.e., tan 26°/cot 64°  = tan 26°/tan 26°  = 1
    

(iii)  cos 48° – sin 42°

Solution :-
  cos 48° – sin 42°
= cos(90° - 42°) - sin 42°
= sin 42° - sin 42°
= 0
    

(iv)  cosec 31° – sec 59°

Solution :-
  cosec 31° – sec 59°
= cosec(90° - 59°) - sec 59°
= sec 59° - sec 59°
= 0
    

Question-2 :-  Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1  (ii) cos 38° cos 52° – sin 38° sin 52° = 0

Solution :-
(i) tan 48° tan 23° tan 42° tan 67° = 1
  L.H.S
  tan 48°. tan 23° . tan 42° . tan 67°
= tan 48° . tan 42° . tan 23° . tan 67°
= tan 48° . tan(90° - 48°) . tan 23° . tan(90° - 23°)
= tan 48° . cot 48° . tan 23° . cot 23°
= tan 48° . 1/tan 48° . tan 23° . 1/tan 23°
= 1
    
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
  L.H.S
  cos 38° . cos 52° – sin 38° . sin 52°
= cos(90° - 52°) . cos 52° - sin(90° - 52°) . sin 52°
= sin 52° . cos 52° - cos 52° . sin 52° 
= 0
    

Question-3 :-  If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution :-
  We are given that tan 2A = cot (A – 18°)..... (1) 
  Since tan 2A = cot (90° – 2A), 
  we can write (1) as cot (90° – 2A) = cot (A – 18°) 
  Since 90° – 2A and A – 18° are both acute angles, 
  Therefore, 
  90° – 2A = A – 18° 
  A + 2A = 90° + 18°
  3A = 108°
   A = 108°/3           
  which gives A = 36°
    

Question-4 :-  If tan A = cot B, prove that A + B = 90°.

Solution :-
  Given :
  tan A = cot B
  we Know that, tan(90° - B) = cot B
  tan A = tan(90° - B)
      A = 90° - B
  A + B = 90° 
    

Question-5 :-  If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Solution :-
  We are given that sec 4A = cosec (A – 20°)..... (1) 
  Since sec 3A = cosec (90° – 4A), 
  we can write (1) as cosec (90° – 4A) = cosec (A – 20°) 
  Since 90° – 4A and A – 20° are both acute angles, 
  Therefore, 
  90° – 4A = A – 20° 
  A + 4A = 90° + 20°
  5A = 110°
   A = 110°/5           
  which gives A = 22°
    

Question-6 :-  If A, B and C are interior angles of a triangle ABC, then show that sin (B+C)/2 = cos A/2.

Solution :-
  We know that sum of angle a triangle = 180°
  A  + B + C = 180
       B + C = 180 - A
       (B + C)/2 = 90 - A/2
  sin (B + C)/2 = sin (90 - A/2)
  sin (B + C)/2 = cos A/2
    

Question-7 :-  Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution :-
  sin 67° + cos 75° 
= sin (90° – 23°) + cos (90° – 15°) 
= cos 23° + sin 15°
    
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