Question

Prove that ${\cot }^{2}A-{\cot }^{2}B=\frac{{\cos }^{2}A-{\cos }^{2}B}{{\sin }^{2}A{\sin }^{2}B}$

Answer 1

L.H.S$={\cot }^{2}A-{\cot }^{2}B$

$=\frac{{\cos }^{2}A}{{\sin }^{2}A}-\frac{{\cos }^{2}B}{{\sin }^{2}B}$

$=\frac{{\cos }^{2}A{\sin }^{2}B-{\cos }^{2}B{\sin }^{2}A}{{\sin }^{2}A{\sin }^{2}B}$

$=\frac{{\cos }^{2}A(1-{\cos }^{2}B)-{\cos }^{2}B(1-{\cos }^{2}A)}{{\sin }^{2}A{\sin }^{2}B}$

$=\frac{{\cos }^{2}A-{\cos }^{2}A{\cos }^{2}B-{\cos }^{2}B+{\cos }^{2}A{\cos }^{2}B}{{\sin }^{2}A{\cos }^{2}B}$

$=\frac{{\cos }^{2}A-{\cos }^{2}B}{{\sin }^{2}A{\sin }^{2}B}$

$=$R.H.S

Hence proved.