1.   Evaluate :
	(i) ${sin 18°}/{cos 72°}$   (ii) ${tan 26° }/{cot 64°}$    (iii)  cos 48° – sin 42°   (iv)  cosec 31° – sec 59°

Solution
Note: General concept for solving these type of question 
$sin θ = cos(90-θ)$ or $cos θ = sin(90-θ)$ and other trigonometric ratios.
(i) ${sin 18°}/{cos 72°}$ = ${sin 18°}/{sin (90° - 72°)}$
   = ${sin 18°}/{sin 18°}$
   =  1

(ii) ${tan 26° }/{cot 64°}$
 = ${tan 26° }/{tan(90-64°)}$ 
 = ${tan 26° }/{tan(26°)}$
= 1 

(iii)  cos 48° – sin 42° 
      = cos 48° – cos(90°- 42°)
      = cos 48° – cos48°
	  = 0 
	  
(iv) cosec 31° – sec 59°
    = cosec 31° – cosec(90- 59°)
    = cosec 31° – cosec 31°
	= 0 
	

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° =  tan 48° tan 23° cot(90° - 42°) cot (90° - 67°)
= tan 48° tan 23°cot48°cot23°
Since $tanθcotθ=1 $
therofore , tan 48° tan 23°cot48°cot23° = 1


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

Solution
Since , we know that tan(θ) =cot(90°-θ),
therefore, tan(2A) =cot(90° -2A)
given, tan 2A = cot (A – 18°)
i.e (90°-2A) = (A-18°)
     3A = 90+ 18
     3A = 108 
	 A = $108/3$ = 36°

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

Solution
Since we know that tan(A) =cot(90°-A) 
and given that tan A = cot B 
               i.e B = 90°-A
               hence, A+B = 90°
 			   

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

Solution
We know that sec θ = cosec(90-θ)
Therefore sec 4A = cosec(90-4A)
Given sec 4A = cosec  (A – 20°)
    therefore (A-20) = 90 -4A 
               5A = 110
                A = 22°			   


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

Solution
Given, A,B, C are the interior of the triangle ABC,
i.e A+B+C = 180° => (B+C) = 180°-A
Therefore sin$({B+C}/2)$ = sin$({180-A}/2)$
                         = sin$(90-A/2)$
			= cos $A/2$
Hence L.H.S = R.H.S
	
7.   Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution
Since sinA= cos(90-A) and cosA = sin(90-A)
Hence sin 67° + cos 75° = cos(90-67°) +sin(90-75°)
                        = cos 23° +sin 15°