$AB^2  + BC^2  =  AC^2  ..................(1)$
		
		Dividing each term of (1) by $AC^2$, we get
		
		${AB^2}/{AC^2} + {BC^2}/{AC^2} = 1$	
		
		$({AB}/{AC})^2 + ({BC}/{AC})^2 = 1$
		
	   i.e., $(cos A)^2  + (sin A)^2  =  1$
	   
	   i.e., $cos^2  A + sin^2  A =  1 $ ................     (2)
	   
	   This is true for all A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity. Let us now divide (1) by $AB^2$. We get
	   Let us now divide (1) by $AB^2$. We get
	   $({AB}/{AB})^2  + ({BC}/{AB})^2  =  ({AC}/{AB})^2$
	   
	   i.e.,  $1 + tan^2 A =  sec^2 A $ .................(3) 

Similarly, on dividing (1) by $BC^2$.
       
	   i.e.,    $cot^2 A + 1 =  cosec^2   A$   ............(4) 

Note that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for all A such that 0° < A ≤ 90°.
Using these identities, we can express each trigonometric ratio in terms of other trigonometric ratios, i.e., 
if any one of the ratios is known, we can also determine the values of other trigonometric ratios.
Let us see how we can do this using these identities. Suppose we know that

	tan A =$1/√3$ then cot A=$√3$
	Since, sec2  A = 1 + tan2  A =  $1+1/3$ =$4/3$ ;
	 therefore; sec A= $2/√3$ and cos A=$√3/2$
	 Again, sin A =$√(1− cos^2 A  )$=$√(1-3/4)$=$1/2$; Therefore, cosec A = 2
	 
	 
Example: Express the ratios cos A, tan A and sec A in terms of sin A.
Solution : Since cos2  A + sin2  A =  1, therefore,
			cos2  A =  1 – sin2  A, i.e.,$ cos A = ±√(1 − sin^2  A)$ 
			Hence,             tan A =${sinA}/{cosA}$ =${sinA}/{√(1 − sin^2  A)}$
			and sec A= $1/{cos A}$ =$1/{√(1 − sin^2  A)}$
			
Example: Prove that sec A (1 – sin A)(sec A + tan A) = 1.
Solution : 
		LHS = sec A (1 – sin A)(sec A + tan A) = $1/{cosA}(1 – sin A)({1}/{cosA}+{sinA}/{cosA})$
												= ${(1-sinA)(1+sinA)}/{cos^2A}$
												=${1-sin^2A}/{cos^2A}$
												=${cos^2A}/{cos^2A}$
												=1
														
Example:Prove that  ${cot A – cos A }/{cot A + cos A}$ =${cosec A – 1}/{cosec A + 1}$  
Solution :  
                LHS =${cot A – cos A }/{cot A + cos A}$ = $ {(cosA)/{sinA} – cos A }/{(cosA)/{sinA} + cos A}$  
				=${cosA({1}/{sinA} – 1 })/{cosA({1}/{sinA} + 1 }$ 
				=${cosecA-1}/{cosecA+1}$ =RHS


Example: Prove  that  ${sin θ − cos θ + 1 }/{sin θ + cos θ − 1}$=$1/{sec θ − tan θ}$ using  the  identity 

				sec2  θ = 1 + tan2  θ.
Solution: Since we will apply the identity involving sec θ  and tan θ, let us first convert the LHS 
                    (of the identity we need to prove) in terms of sec θ and tan θ by dividing numerator and denominator by cos θ.
			LHS= ${sin θ – cos θ + 1 }/{sin θ + cos θ – 1}$
			Divide by cos θ ;
			=${tan θ − 1 + sec θ}/{tan θ + 1 −sec θ}$
			=${((tan θ + sec θ) − 1) (tan θ − sec θ) }/{(tan θ − sec θ + 1) (tan θ− sec θ)}$
			=${((tan^2 θ - sec^2 θ) − (tan θ − sec θ))  }/{(tan θ − sec θ + 1) (tan θ− sec θ)}$
			=${(-1 − tan θ + sec θ)  }/{(tan θ − sec θ + 1) (tan θ− sec θ)}$
			=${-1 }/{ (tan θ− sec θ)}$
			=${1 }/{ (sec θ-tan θ)}$ =RHS