Since ∠ A + ∠ C = 90°, they form complementary angles. Therefore:

	sin A=${BC}/{AC}$ ;   sin C = ${AB}/{AC}$
	cos A=${AB}/{AC}$ ;   cos C =${BC}/{AC}$
	tan A=${BC}/{AB}$ ;   tan C = ${AB}/{BC}$


	Since ∠ A + ∠ C = 90°,
    Therefore,	∠ C = 90° – ∠ A. 
	Hence,
	sin (90° – A) =${AB}/{AC}$ ; 
	cos (90° – A) =${BC}/{AC}$; 
	tan (90° – A) =${AB}/{BC}$
	
	Now, compare the ratios in (1) and (2). Hence :
	sin (90° – A) =${AB}/{AC}$= cos A 	
	
	cos (90° – A) =${BC}/{AC}$ = sin A. 	
	
	tan (90° – A) =${AB}/{BC}$= cot A ;  	
	
	cot (90° – A)=${BC}/{AB}$=tan A. 	
	
	sec (90° – A)=${AC}/{BC}$= cosec A;  	
	
	cosec (90° – A)=${AC}/{AB}$= sec A.	 

Example: Evaluate ${tan 65°}/{cot 25°}$

Solution : cot A =  tan (90° – A)
		   cot 25° =  tan (90° – 25°) = tan 65°
		   i.e ${tan 65°}/{cot 25°}$=${tan 65°}/{tan 65°} = 1$


Example: If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

Solution  We are given that sin 3A = cos (A – 26°).  ............(1)                                           
			Since        sin 3A = cos (90° – 3A), we can write (1) as
			cos (90° – 3A) = cos (A – 26°)
			Since        90° – 3A and A – 26° are both acute angles, therefore,
			90° – 3A =  A – 26°
			which gives   A =  29°
			
Example: Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution :     cot 85° + cos 75° =  cot (90° – 5°) + cos (90° – 15°)
				=  tan 5° + sin 15°