Prove the following identites.
(i) \(\frac{sin \ A - sin \ B}{cos \ A + cos \ B} + \frac{cos \ A - cos \ B}{sin \ A + sin \ B} = 0\)
Proof:
LHS \(= \frac{sin \ A - sin \ B}{cos \ A + cos \ B} + \frac{cos \ A - cos \ B}{sin \ A + sin \ B}\)
\(= \frac{(sin \ A - sin \ B)(sin \ A + sin \ B) + (cos \ A - cos B)(cos \ A + cos \ B)}{(cos \ A + cos \ B)(sin \ A + sin \ B)}\)
\(= \frac{(sin^2 \ A - sin^2 \ B) + (cos^2 \ A - cos^2 \ B)}{(cos \ A + cos \ B)(sin \ A + sin \ B)}\)
By applying the known trigonometric identity \(sin^2 \ A+cos^2 \ A=\) we get,
\(= \frac{(1 - cos^2 \ A) - (1 - cos^2 \ B) + cos^2 \ A - cos^2 \ B}{(cos \ A + cos \ B)(sin \ A + sin \ B)}\)
After simplyfying this then we get R.H.S \(= 0\)
(ii) \(\frac{sin^3 \ A + cos^3 \ A}{sin \ A + cos \ A} + \frac{sin^3 \ A - cos^3 \ A}{sin \ A - cos \ A} = 2\)
Proof:
LHS \(= \frac{sin^3 \ A + cos^3 \ A}{sin \ A + cos \ A} + \frac{sin^3 \ A - cos^3 \ A}{sin \ A - cos \ A}\)
By applying the formula \(a^3+b^3=(\)\()(a^2\)\(+b^2)\) ,
\(a^3+b^3=(\)\()(a^2\)\(+b^2)\) then simplyfing we get,
\(= 2 \ sin^2 \ A + 2 \ cos^2 \ A\)
\(= 2(sin^2 \ A + cos^2 \ A)\)
By applying the known trigonometric identity \(sin^2 \ A+cos^2 \ A=\) , we get the result.